SOLUTION: A ball is thrown straight upward with an initial velocity of 52ft/sec. Its height above the ground after t seconds is given by the formula h(t)=16t^2+52t. At what time will the bal
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Question 595649: A ball is thrown straight upward with an initial velocity of 52ft/sec. Its height above the ground after t seconds is given by the formula h(t)=16t^2+52t. At what time will the ball hit the ground? (in seconds) Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! A ball is thrown straight upward with an initial velocity of 52ft/sec. Its height above the ground after t seconds is given by the formula h(t)=-16t^2+52t. At what time will the ball hit the ground? (in seconds)
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Since height is zero when it hits the ground, set h(t) to zero and solve for t:
h(t)=16t^2+52t
0=-16t^2+52t
0=16t^2-52t
0=t(16t-52)
t = {0, 52/16}
t = {0, 3.25}
answer: 3.25 seconds
