Question

You throw a tennis ball straight up. Air resistance can be neglected.
(a) The maximum height above your hand that the ball reaches is H. At what speed does the ball leave your hand?
(b) What is the speed of the ball when it is a height H/2 above your hand? Express your answer as a fraction of the speed at which it left your hand.
(c) At what height above your hand is the speed of the ball half as great as when it left your hand? Express your answer in terms of H.

Answer 1

Answer:

(a) $\sqrt{2gH}$

(b) $\sqrt{gH}$

(c) $\frac{3u^{2}}{8g}$

Explanation:

The maximum height = H

(a) Let the speed of the ball as it leaves the hand is u.

Use third equation of motion

$v^{2}=u^{2}+2as\\0 = u^{2}- 2 gH\\u=\sqrt{2 gH}$

(b) Let the speed of the ball at H/2 is v.

Use third equation of motion

$v^{2}=u^{2}+2as\\v^{2} = u^{2}- 2 g\frac{H}{2}\\v^{2}=2 g H - gH = gH\\v =\sqrt{gH}$

(c) Let the height is h.

$v^{2}=u^{2}+2as\\\frac{u^{2}}{4} = u^{2}- 2 gh \\\\2 g h = \frac{3u^{2}}{4}\\h =\frac{3u^{2}}{8g}$