You toss a 0.150-kg baseball straight upward so that it leaves your hand moving at 20.0 m/s. The ball reaches a maximum height y2. What is the speed of the ball when it is at a height of y2/2? Ignore air resistance.

14.14 m/s

Explanation:

mass of ball, m 0.150 kg

initial velocity, u = 20 m/s

g = 10 m/s²

The final velocity at the height is zero. Total height is y2.

Use third equation of motion,

v² = u² - 2 gh

0 = 20 x 20 - 2 x 10 x y2

y2 = 20 m

Let v is the velocity at height y2/2.

v² = u² - 2 gh

v² = 20 x 20 - 2 x 10 x 10

v² = 400 - 200

v = 14.14 m/s

Thus, the velocity at half of the height is 14.14 m/s.