Question 118277: A ball is thrown upward from the to of a 98-meter tower with an initial speed of 39.2 m/s. How much later will it hit the ground? Hint: If h is the height of the ball above the top of the tower, then h=98 when the ball hits the ground
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A ball is thrown upward from the to of a 98-meter tower with an initial speed of 39.2 m/s. How much later will it hit the ground? Hint: If h is the height of the ball above the top of the tower, then h=98 when the ball hits the ground.
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Formula: h(t) = -9.81t^2+vot+s(o)
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Your Problem:
h(t) = -9.81t^2+39.2t+98
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Comment: Usually "hit the ground" means h=0, not h=98.
I'm going to assume you mean h=0.
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-9.81t^2+39.2t+98=0
t = [-39.2 +- sqrt(39.2^2-4*-9.81*98)]/[2*-9.81]
t = [-39.2 +- 73.36]/[-19.62]
t = 5.74 seconds
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Cheers,
Stan H.
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