A ball is thrown into the air with an initial upward velocity of 48 ft/s. Its height (h) in feet after t seconds is given by the function h=-16t^2+48t+64. After how many seconds will the ball hit the ground?
Responses
5 seconds
7 seconds
4 seconds
6 seconds
The ball will hit the ground when h=0. So, we need to solve the equation -16t^2+48t+64=0.
Dividing both sides by -16, we get t^2 - 3t - 4 = 0.
Factoring, we get (t-4)(t+1) = 0.
Therefore, the ball will hit the ground after 4 seconds or -1 second (which doesn't make physical sense).
So, the answer is 4 seconds.