Question 656894: If I throw a baseball 7.5 m up in the air, how long will it take to get back to my hand?
I know that delta x is 7.5m, and I know that I need to find t, for time, and I also know that his acceleration is 9.8 meters per second squared. I used the equation delta x= v1(t) + 1/2(a)(t)^2, and I assumed that v1 would be 0, but I'm not right, and I don't understand why.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! If I throw a baseball 7.5 m up in the air, how long will it take to get back to my hand?
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x= v1(t) + 1/2(a)(t)^2,
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If v1 = 0, then the ball wouldn't go up.
v1 is not given.
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The time to get to the apogee is the same time to fall back to the hand.
The height of the hand is not given, so I assume the 7.5 m is from the hand's height.
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The ball falls from 7.5 meters
h(t) = -4.9t^2 + 7.5 = 0
t^2 = 7.5/4.9
t =~ 1.2372 seconds rising, and the same time falling.
Total time =~ 2.4744 seconds
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