SOLUTION: During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 5.91 s, how
high above the point where it hits the bat
does it rise? Assume when it hits
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-> SOLUTION: During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 5.91 s, how
high above the point where it hits the bat
does it rise? Assume when it hits
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Question 1165529: During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 5.91 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat. The
acceleration of gravity is 9.8 m/s^2
.
Answer in units of m. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! the vertex value of time is half of 5.91 s or 2.955 sec
That means it fell from the maximum height and hit the ground in 2.955 sec
s=1/2 (at^2)
=0.5*9.8 m/sec ^2*2.955^2
=42.79 m
