You throw a baseball directly upward at time t=0 at an initial speed of 14.7 m/s
ID: 2102777 • Letter: Y
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