You throw a baseball directly upward at time t=0 at an initial speed of 14.7 m/s

Question

You throw a baseball directly upward at time t=0 at an initial speed of 14.7 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what time does the ball pass through half the maximum height? Ignore air resistance and take g=9.80 m/s^2

Need to know:

Maximum height

Earlier time at half maximum height

Later time at half maximum height

Explanation / Answer

v^2- u^2 = 2a *s => 0 - 14.7^2 = 2*-9.8 *s => s= 11.025 m s= ut+ 0.5 at^2 => 11.025/2 = 14.7 t -0.5*9.8 t^2 => t =0.44 s, 2.56 s a). H = 11.025 s b). 0.44 s c). 2.56 s

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