Question
A ball is thrown straight upward with an initial velocity of 25 m/s.
(a) How high does it rise?
(b) How long does it take to return to the thrower’s hand? (Take g = 9.8 m/s²)
Quick Answer
Initial velocity u = 25 m/s
Final velocity at top v = 0 m/s
Acceleration a = –9.8 m/s²
(a) Maximum height h = (v² – u²) / (2a) = (0² – 25²) / (2 × –9.8) = 625 / 19.6 ≈ 31.9 m
(b) Time to rise = (v – u) / a = (0 – 25) / –9.8 ≈ 2.55 s
Total time = 2 × 2.55 = 5.10 s
The ball rises to a height of about 31.9 metres and takes about 5.1 seconds to return.
Understanding the question with detailed explanations
A ball is projected vertically upward with an initial velocity of 25 metres per second. We are asked two things: first, the maximum height the ball reaches before momentarily coming to rest, and second, the total time taken for the ball to rise and then return to the thrower’s hand. The acceleration due to gravity is given as 9.8 m/s², and it acts downward, opposing the motion of the ball as it rises.
This question is a classic example of uniformly accelerated motion under gravity. When the ball is thrown upward, it gradually slows down until its velocity becomes zero at the maximum height. Then, it accelerates downward due to gravity until it returns to the original level. By applying equations of motion carefully, we can determine both the height and the total time of flight.
The motion of the ball can be split into two stages: the upward journey and the downward journey. On the way up, the ball slows down under the effect of gravity until its velocity becomes zero. At that point, it has reached its maximum height. The distance travelled during this upward motion is what we need for part (a). On the way down, gravity accelerates the ball back toward the ground at the same rate, bringing it back to the thrower’s hand.
The total time of flight is the sum of the time taken to rise and the time taken to fall. Because gravity is constant and uniform, the time taken to go up is equal to the time taken to come down. Thus, once we find the time to rise, we can simply double it to get the total time for part (b). These steps illustrate the symmetry in projectile motion under constant gravity.
A glimpse of the final answer
From the calculations, the ball reaches a maximum height of about 31.9 metres. This value makes sense: with an initial speed of 25 m/s, the ball should travel several tens of metres upward before coming to rest under the pull of gravity. The number is neither too small nor unrealistically large, showing consistency with everyday expectations.
The total time of flight comes out as 5.1 seconds. This also seems logical: about 2.5 seconds to go up, and another 2.5 seconds to come down. The motion is symmetric, so the times are equal. These results confirm that the equations have been applied correctly, and they match the physical situation well.
Data
From the problem:
- Initial velocity, u = 25 m/s
- Final velocity at top, v = 0 m/s
- Acceleration due to gravity, a = –9.8 m/s²
- Displacement at maximum height = ?
- Time of flight = ?
What we need:
- Maximum height reached, h
- Total time to return to the thrower’s hand, T
All values are already given in SI units, so no conversions are required.
Formula
Relevant equations of motion are:
- v² = u² + 2as (to find maximum height)
- v = u + at (to find time to rise)
The first equation is used because it relates velocity, acceleration, and displacement, and at the highest point the velocity is zero. The second equation is used to find the time taken to rise. Since the time to descend is equal, multiplying by 2 gives the total time of flight.
These formulas are perfectly suited here because the acceleration due to gravity is constant and acts uniformly on the ball throughout the motion.
Solution (solving the problem)
(a) Maximum height
v² = u² + 2as
0² = 25² + 2(–9.8)(s)
0 = 625 – 19.6s
19.6s = 625
s = 625 / 19.6 ≈ 31.9 m
(b) Time to rise
v = u + at
0 = 25 + (–9.8)t
–25 = –9.8t
t = 25 / 9.8 ≈ 2.55 s
Total time of flight = 2 × 2.55 ≈ 5.10 s
Final Answer
The ball reaches a maximum height of approximately 31.9 metres above the thrower’s hand.
The total time taken to return to the thrower’s hand is approximately 5.1 seconds.
Helpful Explanation
This problem demonstrates how gravity controls the motion of objects thrown upward. Even though the ball starts fast at 25 m/s, gravity steadily reduces its speed until it momentarily stops at the top. Then, gravity accelerates it back down at the same rate, making the ascent and descent times equal. This symmetry is a hallmark of projectile motion.
A common mistake in such questions is forgetting to treat acceleration as negative when the ball is going upward. This sign convention ensures that velocity decreases with time. Another common error is to forget doubling the rise time to get the total flight time. An exam tip is to always separate the upward and downward journeys, solve for one, and then use the symmetry to get the other.