Question 153158: Please help!!!
A ball is thrown upward with an initial speed of 50ft/sec. It's height after t seconds is given by h=50t - 16t^2.
A) What is the height of the ball after 1.3 seconds?
B) When does the ball hit the ground?
For A, the answer I worked out is 37.96 feet.
For B, I worked out 25/8 seconds = t, which doesn't seem right. Can anyone help?
Thank you!!!!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A ball is thrown upward with an initial speed of 50ft/sec. It's height after t seconds is given by h=50t - 16t^2.
A) What is the height of the ball after 1.3 seconds?
For A, the answer I worked out is 37.96 feet.
That is correct.
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B) When does the ball hit the ground?
Let h = 0.
0 = 50t -16t^2
Factor to get:
2t(25 - 8t) = 0
t = 0 or t = 25/8 sec.
Comment: that tells you the ball is on the ground at the beginning
and is back on the ground after 25/8 = 3 1/8 seconds.
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For B, I worked out 25/8 seconds = t, which doesn't seem right. Can anyone help?
You are correct
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Cheers,
Stan H.
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