A physics student throws a ball straight up. The student catches the ball in exactly the same place from which it was released. The ball’s time of flight is T, and its maximum height above its release point is H. Neglect air resistance and assume up is the positive direction. Find the ball’s average velocity during the second half of its trip. (Hint: Your answer should only have the variables H and T in it.)

The correct answer is H ÷ ¹/₂T

Explanation:

The formula for velocity is distance covered ÷ time.

Neglecting air resistance;

If the ball's time of overall time flight is T, the time it will take for the second half/return trip is ¹/₂T.

If the ball's maximum height above its released point is H, the height will also be the distance it covered for the second part of the trip since the student caught the ball in the exact same place the ball was thrown. Hence, the distance for the second half of the trip will be H.

Since velocity = distance/time

The average velocity during the second half of the trip will be = H ÷ ¹/₂T