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The height, h, in feet of a golf ball above the ground after being hit into the air is given by the equation, h = -16t^2 + 64t, where t is the number of seconds elapsed since the ball was hit. How many seconds does it take for the golf ball to hit the ground?

Verified Answer :

LammettHash

The ball is on the ground whenever the height is 0:

\(-16t^2+64t=0\implies-16t(t-4)=0\implies t=0,t=4\)

Obviously, \(t=0\) is the time when the ball is thrown, so ignore that. It thus takes the ball \(t=4\) seconds to hit the ground after it's been launched.

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