You throw a baseball directly upward at time t = 0 at an initial speed of 13.9 m

Question

You throw a baseball directly upward at time t = 0 at an initial speed of 13.9 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/s2.

You throw a baseball directly upward at time t = 0 at an initial speed of 13.9 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/s2. Maximum height: Earlier time at half maximum height: Later time at half maximum height:

Explanation / Answer

at max ht, velocity = 0, so using the SUVAT equation
v

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