When a body is thrown vertically upwards, its velocity at the maximum height is

Analyze the question A body is thrown straight up. We are asked about its velocity at the very top of its path (maximum height).

Concept Gravity acts downwards with acceleration g ≈ 10 m/s² (or 9.8 m/s²). As the body rises, gravity opposes its motion, so it slows down steadily. At the highest point, it momentarily stops before it begins to fall back down.

Solution Use the equation of motion:

v = u − gt

At maximum height, the body stops rising, so v = 0.

Check with another equation:

v² = u² − 2gh 0 = u² − 2gH H = u²/(2g)

Here, the velocity is 0 at the height H. For example, if u = 20 m/s, the time to reach the top is t = u/g = 20/10 = 2 s, and v = 20 − 10(2) = 0.

Checking the options

• A. Maximum: wrong. The velocity is greatest at the start (when thrown), not at the top.
• B. Zero: correct. The body momentarily comes to rest.
• C. Double its initial value: wrong. The body is decelerating, so it can't gain speed.
• D. Half its initial value: wrong. The velocity doesn't drop only to half. It drops all the way to zero.

Exam tip Remember: at maximum height, v = 0, but acceleration is not zero. Gravity still acts at g downwards. WASSCE and JAMB often test this distinction. Also, the time going up equals the time coming down (ignoring air resistance), so total time of flight is T = 2u/g.