You throw a baseball directly upward at time t = 0 at an initial speed of 12.9 m

Question

You throw a baseball directly upward at time t = 0 at an initial speed of 12.9 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.

Hint*

How do the initial speed and the acceleration due to gravity relate to the time the ball reaches maximum height? How do this time, the initial speed, and the acceleration due to gravity relate to the maximum height? How do the maximum height, the initial speed, and the acceleration due to gravity relate to the times at which the ball passes through half the maximum height?

Explanation / Answer

H = u^2 / 2g

H = (12.9)^2 / 2*9.8

H = 8.49 m

h = H/2 = 4.245 m

h = ut + at^2 / 2

4.245 = 12.9t - 9.8t^2 / 2

4.9t^2 - 12.9t + 4.245 = 0

t = 0.3855 sec. (and) t = 2.247 sec.

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