You throw a baseball directly upward at time f 0 at an init

You throw a baseball directly upward at time f = 0 at an initial speed of 13.3 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s^2.

Solution

Herem

initial speed , u = 13.3 m/s

Now , for the maximum height

maximum height = u^2/(2 * g)

maximum height = 13.3^2/(2 * 9.8)

maximum height = 9.03 m
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let the time of half maximum height is t

h = u * t - 0.5 * g * t^2

9.03/2 = 13.3 * t - 0.5 * 9.8 * t^2

solving for t

t = 0.397 s , 2.32 s

the earlier time is 0.397 s

the later time for half maximum height is 2.32 s