301. A stone is thrown vertically upward and returns to the same level after \(4\,\text{s}\). Neglecting air resistance, the time taken to reach the highest point is
ⓐ. \(1\,\text{s}\)
ⓑ. \(2\,\text{s}\)
ⓒ. \(3\,\text{s}\)
ⓓ. \(4\,\text{s}\)
Correct Answer: \(2\,\text{s}\)
Explanation: For vertical motion under gravity without air resistance, the upward and downward parts are symmetric when the body returns to the same level. The total time of flight is \(4\,\text{s}\). The highest point occurs halfway through the motion because the time of ascent equals the time of descent. Therefore,
\[
t_{\text{up}}=\frac{4\,\text{s}}{2}
\]
\[
t_{\text{up}}=2\,\text{s}
\]
The symmetry applies because the launch and return levels are the same and acceleration due to gravity is constant.
\( \textbf{Final answer:} \) The time to reach the highest point is \(2\,\text{s}\).
302. A ball is thrown upward and returns to the same point with no air resistance. Just before returning to the starting point, its velocity is
ⓐ. equal in magnitude and same in direction as the initial velocity
ⓑ. zero because it returns to the starting point
ⓒ. greater than the initial velocity because gravity adds speed both ways
ⓓ. equal in magnitude and opposite in direction to the initial velocity
Correct Answer: equal in magnitude and opposite in direction to the initial velocity
Explanation: When a ball returns to the same height in motion under gravity without air resistance, mechanical symmetry gives equal speed at the same level. On the way up, the velocity is upward. On the way down at the same level, the velocity is downward. Thus the magnitude is the same but the direction is opposite. If upward is taken positive, the return velocity is negative while the initial velocity was positive. Returning to the starting point does not mean the velocity is zero.
303. A stone is thrown vertically downward with \(u=5\,\text{m s}^{-1}\) from the top of a building. Taking downward as positive and \(g=10\,\text{m s}^{-2}\), its velocity after \(2\,\text{s}\) is
ⓐ. \(+10\,\text{m s}^{-1}\)
ⓑ. \(+15\,\text{m s}^{-1}\)
ⓒ. \(+25\,\text{m s}^{-1}\)
ⓓ. \(-25\,\text{m s}^{-1}\)
Correct Answer: \(+25\,\text{m s}^{-1}\)
Explanation: \( \textbf{Sign convention:} \) Downward is positive.
\( \textbf{Initial velocity:} \) \(u=+5\,\text{m s}^{-1}\).
\( \textbf{Acceleration:} \) \(a=+10\,\text{m s}^{-2}\).
\( \textbf{Time:} \) \(t=2\,\text{s}\).
Use:
\[
v=u+at
\]
Substitute:
\[
v=5+(10)(2)
\]
\[
v=5+20
\]
\[
v=+25\,\text{m s}^{-1}
\]
The positive sign means the velocity is downward in the chosen convention.
\( \textbf{Final answer:} \) The velocity after \(2\,\text{s}\) is \(+25\,\text{m s}^{-1}\).
304. A freely falling body starts from rest. Neglecting air resistance, the distances covered in the \(1^\text{st}\), \(2^\text{nd}\), and \(3^\text{rd}\) seconds are in the ratio
ⓐ. \(1:2:3\)
ⓑ. \(1:3:5\)
ⓒ. \(1:4:9\)
ⓓ. \(2:4:6\)
Correct Answer: \(1:3:5\)
Explanation: A freely falling body starting from rest has \(u=0\,\text{m s}^{-1}\) and constant acceleration \(g\). For uniformly accelerated motion from rest, displacement in the \(n^\text{th}\) second is proportional to \(2n-1\). Therefore, the successive one-second distances are proportional to:
\[
1,\quad3,\quad5,\quad7,\ldots
\]
For the first three seconds, the ratio is:
\[
1:3:5
\]
The total distances from the start after \(1\,\text{s}\), \(2\,\text{s}\), and \(3\,\text{s}\) would instead be in the ratio \(1:4:9\).
\( \textbf{Final answer:} \) The distances in successive seconds are in the ratio \(1:3:5\).
305. A ball is thrown vertically upward with \(u=30\,\text{m s}^{-1}\). Taking upward as positive and \(g=10\,\text{m s}^{-2}\), its velocity after \(3\,\text{s}\) is
ⓐ. \(+30\,\text{m s}^{-1}\)
ⓑ. \(+15\,\text{m s}^{-1}\)
ⓒ. \(0\,\text{m s}^{-1}\)
ⓓ. \(-30\,\text{m s}^{-1}\)
Correct Answer: \(0\,\text{m s}^{-1}\)
Explanation: \( \textbf{Sign convention:} \) Upward is positive.
\( \textbf{Initial velocity:} \) \(u=+30\,\text{m s}^{-1}\).
\( \textbf{Acceleration:} \) Gravity acts downward, so \(a=-10\,\text{m s}^{-2}\).
\( \textbf{Time:} \) \(t=3\,\text{s}\).
Use:
\[
v=u+at
\]
Substitute:
\[
v=30+(-10)(3)
\]
\[
v=30-30
\]
\[
v=0\,\text{m s}^{-1}
\]
The ball is momentarily at the highest point at this instant, while its acceleration is still downward.
\( \textbf{Final answer:} \) The velocity after \(3\,\text{s}\) is \(0\,\text{m s}^{-1}\).
306. A ball is thrown vertically upward with \(u=30\,\text{m s}^{-1}\). Taking upward as positive and \(g=10\,\text{m s}^{-2}\), its height above the point of projection after \(3\,\text{s}\) is
ⓐ. \(30\,\text{m}\)
ⓑ. \(45\,\text{m}\)
ⓒ. \(60\,\text{m}\)
ⓓ. \(90\,\text{m}\)
Correct Answer: \(45\,\text{m}\)
Explanation: \( \textbf{Initial velocity:} \) \(u=+30\,\text{m s}^{-1}\).
\( \textbf{Acceleration:} \) Since upward is positive, \(a=-10\,\text{m s}^{-2}\).
\( \textbf{Time:} \) \(t=3\,\text{s}\).
Use:
\[
s=ut+\frac{1}{2}at^2
\]
Substitute:
\[
s=(30)(3)+\frac{1}{2}(-10)(3^2)
\]
\[
s=90-5(9)
\]
\[
s=90-45
\]
\[
s=45\,\text{m}
\]
This is also the maximum height because the velocity becomes zero at \(t=3\,\text{s}\).
\( \textbf{Final answer:} \) The height above the point of projection is \(45\,\text{m}\).
307. A ball is thrown vertically upward with speed \(30\,\text{m s}^{-1}\) and returns to the same level. Taking \(g=10\,\text{m s}^{-2}\), the total time of flight is
ⓐ. \(3\,\text{s}\)
ⓑ. \(4\,\text{s}\)
ⓒ. \(6\,\text{s}\)
ⓓ. \(9\,\text{s}\)
Correct Answer: \(6\,\text{s}\)
Explanation: \( \textbf{Initial velocity upward:} \) \(u=30\,\text{m s}^{-1}\).
For upward motion, time to reach the highest point is found from:
\[
v=u-gt
\]
At the highest point, \(v=0\,\text{m s}^{-1}\):
\[
0=30-10t
\]
\[
t=3\,\text{s}
\]
When the ball returns to the same level and air resistance is neglected, time of descent equals time of ascent.
Therefore:
\[
T=2t=2(3\,\text{s})
\]
\[
T=6\,\text{s}
\]
The symmetry is valid because the initial and final levels are the same.
\( \textbf{Final answer:} \) The total time of flight is \(6\,\text{s}\).
308. A ball is thrown vertically upward with velocity \(+u\) and later returns to the same level without air resistance. Taking upward as positive, the velocity just before reaching the starting level is
ⓐ. \(+u\)
ⓑ. \(-u\)
ⓒ. \(0\)
ⓓ. \(+2u\)
Correct Answer: \(-u\)
Explanation: At the same height, the speed of the ball is the same in upward and downward motion when air resistance is neglected. The direction, however, changes. Initially the ball has upward velocity \(+u\). Just before it returns to the starting level, it is moving downward, which is negative in the chosen sign convention. Therefore, the velocity is \(-u\), not \(+u\). The equality is in speed magnitude, while the sign records the opposite direction.
309. Match each vertical-motion situation with the correct sign description when upward is chosen as positive.
| Situation | Sign description |
|---|
| P. Ball moving upward | 1. \(v\gt0\) |
| Q. Ball moving downward | 2. \(v\lt0\) |
| R. Acceleration due to gravity | 3. \(a=-g\) |
| S. Highest point | 4. \(v=0\), \(a=-g\) |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-1, Q-3, R-2, S-4
ⓒ. P-4, Q-2, R-3, S-1
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: If upward is chosen as positive, upward velocity is positive and downward velocity is negative. Gravity acts downward, so acceleration due to gravity is \(a=-g\). At the highest point, the ball is momentarily at rest, so \(v=0\), but gravity still acts downward. Therefore, the acceleration remains \(-g\) even at the top. The highest point is not a point of zero acceleration in free vertical motion.
310. A stone is dropped from rest from a height of \(80\,\text{m}\). Taking downward as positive and \(g=10\,\text{m s}^{-2}\), the time taken to reach the ground is
ⓐ. \(2\,\text{s}\)
ⓑ. \(4\,\text{s}\)
ⓒ. \(6\,\text{s}\)
ⓓ. \(8\,\text{s}\)
Correct Answer: \(4\,\text{s}\)
Explanation: \( \textbf{Initial velocity:} \) Dropped from rest gives \(u=0\,\text{m s}^{-1}\).
\( \textbf{Displacement:} \) Downward displacement is \(s=+80\,\text{m}\).
\( \textbf{Acceleration:} \) \(a=+10\,\text{m s}^{-2}\).
Use:
\[
s=ut+\frac{1}{2}at^2
\]
Substitute:
\[
80=0+\frac{1}{2}(10)t^2
\]
\[
80=5t^2
\]
\[
t^2=16
\]
\[
t=4\,\text{s}
\]
Only the positive value of time is physically meaningful.
\( \textbf{Final answer:} \) The stone reaches the ground in \(4\,\text{s}\).
311. A stone is thrown vertically downward with \(u=10\,\text{m s}^{-1}\) from a height of \(15\,\text{m}\). Taking downward as positive and \(g=10\,\text{m s}^{-2}\), the time taken to hit the ground is
ⓐ. \(0.5\,\text{s}\)
ⓑ. \(1.0\,\text{s}\)
ⓒ. \(1.5\,\text{s}\)
ⓓ. \(2.0\,\text{s}\)
Correct Answer: \(1.0\,\text{s}\)
Explanation: \( \textbf{Sign convention:} \) Downward is positive.
\( \textbf{Initial velocity:} \) \(u=+10\,\text{m s}^{-1}\).
\( \textbf{Displacement:} \) \(s=+15\,\text{m}\).
\( \textbf{Acceleration:} \) \(a=+10\,\text{m s}^{-2}\).
Use:
\[
s=ut+\frac{1}{2}at^2
\]
Substitute:
\[
15=10t+\frac{1}{2}(10)t^2
\]
\[
15=10t+5t^2
\]
Check \(t=1\,\text{s}\):
\[
10(1)+5(1)^2=15
\]
So the equation is satisfied.
The downward throw already gives the stone an initial downward velocity, so the time is less than it would be for a drop from rest from the same height.
\( \textbf{Final answer:} \) The time taken is \(1.0\,\text{s}\).
312. A stone is dropped from rest. If it falls \(5\,\text{m}\) in the \(1^\text{st}\) second, then the distance fallen in the \(4^\text{th}\) second is
ⓐ. \(15\,\text{m}\)
ⓑ. \(25\,\text{m}\)
ⓒ. \(35\,\text{m}\)
ⓓ. \(45\,\text{m}\)
Correct Answer: \(35\,\text{m}\)
Explanation: A dropped stone starts from rest, so its motion has \(u=0\,\text{m s}^{-1}\) and constant acceleration \(g\). For uniform acceleration from rest, distances in successive seconds are in the ratio:
\[
1:3:5:7:\cdots
\]
The \(1^\text{st}\)-second distance corresponds to \(1\) part.
The \(4^\text{th}\)-second distance corresponds to \(7\) parts.
Given:
\[
1\text{ part}=5\,\text{m}
\]
Therefore:
\[
7\text{ parts}=35\,\text{m}
\]
This is the distance during the \(4^\text{th}\) second only, not the total distance in \(4\,\text{s}\).
\( \textbf{Final answer:} \) The distance fallen in the \(4^\text{th}\) second is \(35\,\text{m}\).
313. A student claims, “At the highest point of vertical motion, both velocity and acceleration are zero.” The correct correction is that
ⓐ. zero acceleration, but maximum velocity
ⓑ. maximum velocity and maximum acceleration
ⓒ. downward velocity, but upward acceleration
ⓓ. zero velocity, but acceleration \(g\) downward
Correct Answer: zero velocity, but acceleration \(g\) downward
Explanation: At the highest point, the ball is momentarily changing direction, so its instantaneous velocity is \(0\,\text{m s}^{-1}\). Gravity does not stop acting at that point. The acceleration is still directed downward with magnitude \(g\). If upward is chosen positive, this acceleration is written as \(-g\). Confusing zero velocity with zero acceleration is a common error because a body can have zero velocity at an instant while still accelerating.
314. The \(v-t\) graph of a ball thrown vertically upward is a straight line with negative slope when upward is positive. What does the negative slope represent?
ⓐ. Displacement negative throughout the motion
ⓑ. Velocity zero throughout the motion
ⓒ. Speed increasing upward throughout the motion
ⓓ. Acceleration downward due to gravity
Correct Answer: Acceleration downward due to gravity
Explanation: On a \(v-t\) graph, the slope represents acceleration. For a ball thrown upward, gravity acts downward throughout the motion. If upward is taken as positive, downward acceleration is negative, so the \(v-t\) graph has a negative slope. The graph may cross the time axis when the velocity becomes zero at the highest point. The negative slope does not mean the ball is always moving downward; it means velocity is decreasing with time in the chosen positive direction.
315. A ball is thrown vertically upward with \(u=10\,\text{m s}^{-1}\). Taking upward as positive and \(g=10\,\text{m s}^{-2}\), its displacement after \(2\,\text{s}\) is
ⓐ. \(5\,\text{m}\)
ⓑ. \(10\,\text{m}\)
ⓒ. \(0\,\text{m}\)
ⓓ. \(20\,\text{m}\)
Correct Answer: \(0\,\text{m}\)
Explanation: \( \textbf{Initial velocity:} \) \(u=+10\,\text{m s}^{-1}\).
\( \textbf{Acceleration:} \) \(a=-10\,\text{m s}^{-2}\).
\( \textbf{Time:} \) \(t=2\,\text{s}\).
Use:
\[
s=ut+\frac{1}{2}at^2
\]
Substitute:
\[
s=(10)(2)+\frac{1}{2}(-10)(2^2)
\]
\[
s=20-5(4)
\]
\[
s=20-20
\]
\[
s=0\,\text{m}
\]
The ball has returned to the point of projection after \(2\,\text{s}\).
Zero displacement here does not mean the ball did not move; it went up and came back.
\( \textbf{Final answer:} \) The displacement after \(2\,\text{s}\) is \(0\,\text{m}\).
316. A ball is thrown vertically upward with \(u=10\,\text{m s}^{-1}\). Taking \(g=10\,\text{m s}^{-2}\), the total distance travelled in the first \(2\,\text{s}\) is
ⓐ. \(0\,\text{m}\)
ⓑ. \(5\,\text{m}\)
ⓒ. \(10\,\text{m}\)
ⓓ. \(20\,\text{m}\)
Correct Answer: \(10\,\text{m}\)
Explanation: \( \textbf{Maximum height:} \) At the top, \(v=0\,\text{m s}^{-1}\).
Use:
\[
v^2=u^2+2as
\]
With \(u=10\,\text{m s}^{-1}\) and \(a=-10\,\text{m s}^{-2}\):
\[
0^2=10^2+2(-10)s
\]
\[
0=100-20s
\]
\[
s=5\,\text{m}
\]
The ball travels \(5\,\text{m}\) upward and then \(5\,\text{m}\) downward back to the starting point in \(2\,\text{s}\).
Thus distance travelled is:
\[
5\,\text{m}+5\,\text{m}=10\,\text{m}
\]
The displacement is zero, but distance counts both parts of the path.
\( \textbf{Final answer:} \) The total distance travelled is \(10\,\text{m}\).
317. Two stones are released from rest from the same height, one after the other, neglecting air resistance. Once both are in air, their relative acceleration is
ⓐ. \(g\) downward
ⓑ. \(2g\) downward
ⓒ. \(g\) upward
ⓓ. \(0\)
Correct Answer: \(0\)
Explanation: Both stones are freely falling under gravity alone. Each has the same downward acceleration \(g\). Relative acceleration is the difference between their accelerations. Since both accelerations are equal, their relative acceleration is zero. This does not mean their relative velocity is necessarily zero; if one stone was released earlier, it may already have a larger downward velocity.
318. A car \(P\) moves east with velocity \(20\,\text{m s}^{-1}\), and car \(Q\) moves east with velocity \(12\,\text{m s}^{-1}\). The velocity of \(P\) relative to \(Q\) is
ⓐ. \(12\,\text{m s}^{-1}\) east
ⓑ. \(20\,\text{m s}^{-1}\) east
ⓒ. \(32\,\text{m s}^{-1}\) east
ⓓ. \(8\,\text{m s}^{-1}\) east
Correct Answer: \(8\,\text{m s}^{-1}\) east
Explanation: \( \textbf{Choose east as positive:} \)
\[
v_P=+20\,\text{m s}^{-1}
\]
\[
v_Q=+12\,\text{m s}^{-1}
\]
Velocity of \(P\) relative to \(Q\) is:
\[
v_{PQ}=v_P-v_Q
\]
Substitute:
\[
v_{PQ}=20-12
\]
\[
v_{PQ}=+8\,\text{m s}^{-1}
\]
The positive sign means \(P\) appears to move east relative to \(Q\).
Relative velocity compares velocities by subtraction, not by simply naming the faster car.
\( \textbf{Final answer:} \) The velocity of \(P\) relative to \(Q\) is \(8\,\text{m s}^{-1}\) east.
319. Two cars move in opposite directions on a straight road. Car \(P\) moves east at \(15\,\text{m s}^{-1}\), and car \(Q\) moves west at \(10\,\text{m s}^{-1}\). Taking east as positive, the velocity of \(P\) relative to \(Q\) is
ⓐ. \(+5\,\text{m s}^{-1}\)
ⓑ. \(-5\,\text{m s}^{-1}\)
ⓒ. \(+25\,\text{m s}^{-1}\)
ⓓ. \(-25\,\text{m s}^{-1}\)
Correct Answer: \(+25\,\text{m s}^{-1}\)
Explanation: \( \textbf{Sign convention:} \) East is positive.
Car \(P\) has:
\[
v_P=+15\,\text{m s}^{-1}
\]
Car \(Q\) moves west, so:
\[
v_Q=-10\,\text{m s}^{-1}
\]
Relative velocity of \(P\) with respect to \(Q\) is:
\[
v_{PQ}=v_P-v_Q
\]
\[
v_{PQ}=+15-(-10)
\]
\[
v_{PQ}=+25\,\text{m s}^{-1}
\]
The positive sign shows that \(P\) appears to move east relative to \(Q\).
\( \textbf{Final answer:} \) \(v_{PQ}=+25\,\text{m s}^{-1}\).
320. If two bodies move along the same straight line with velocities \(v_A\) and \(v_B\), the relative velocity of \(A\) with respect to \(B\) is
ⓐ. \(v_A+v_B\) always
ⓑ. \(v_A-v_B\)
ⓒ. \(v_B-v_A\) always
ⓓ. \(\frac{v_A}{v_B}\)
Correct Answer: \(v_A-v_B\)
Explanation: Relative velocity of \(A\) with respect to \(B\) means the velocity of \(A\) as observed from \(B\). In one-dimensional motion, it is found by subtracting the velocity of the observer body from the velocity of the observed body. Thus:
\[
v_{AB}=v_A-v_B
\]
The signs of \(v_A\) and \(v_B\) must be written according to the same coordinate convention before subtraction. Adding speeds only works as a special-looking result in some opposite-direction cases, not as the general rule.
