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

You throw a baseball directly upward at time t = 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

ball is thrown vertically upwards with a speed of 13.3 m/s

maximum height the ball reaches

mgh = 0.5mv^2

h = v^2/2g

h(max) =13.3^2 / 2*9.8 = 9.025 m

half height = 9.025/2 = 4.5125 m

v^2-u^2 = 2(-g)S

v^2 - 13.3^2 = 2*-9.8*4.5125

v = 9.4 m/s

v = u - gt

9.4 = 13.3 - 9.8*t

t = 0.397 s

time taken to reach to maximum point = 2*t(half) = 2*0.397 = 0.794 s

time taken to last time half maximu height = 3*0.794= 1.91 s