You are watching a softball game. The batter hits a pop-fly such that the ball g

Question

You are watching a softball game. The batter hits a pop-fly such that the ball goes straight up with a velocity of 10.0 m/s. The bat made contact with the ball exactly 1.00 m above the ground.

How high above the ground will the ball go? (what is the ball's maximum height)

How much time does the catcher have to catch the ball, assuming she catches the ball 1.00 m above the ground (same height as where the ball was hit?)

The catcher misses the ball and the ball hits the ground. What is the ball's velocity when it hits the gorund?

Explanation / Answer

here,

initial velocity , u = 10 m/s

the height of ball above the ground , h = 1 +u^2/2g

h = 1 + 10^2 /( 2 * 9.81)

h = 6.1 m

the time does the catcher have , t = 2 * u/g

t = 2 * 10 /9.81 = 2.04 s

let the final speed be v

v^2 - u^2 = 2 * g* 1

v^2 = 10^2 + 2 * 9.81 * 1

v = 10.9 m/s

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