A ball is launched upward from an initial height of 160 feet

A ball is launched upward from an initial height of 160 feet off the ground. The height (in feet) of the ball t seconds after it is thrown is given by the function h (t) = —16t^2+144t+160 When does the ball reach the highest point?

Solution

We know that h(t) is maximum when dh/dt = 0 and d²h/dt² is negative. Here, dh/dt = -32t +144 and d²h/dt² = -32. Thus dh/dt = 0 when -32t +144 = 0 or, when t = 144/32 = 4.5 seconds. Further, d²h/dt² is always negative regardless of the value of t. Therefore, the ball takes 4.5 seconds in reaching the highest point.

Alternate Method:

A ball is launched upward from an initial height of 160 feet off the ground. The height (in feet) of the ball t seconds after it is thrown is given by the function h (t) = -16t²+144t+160 = -16(t² – 9t) +160 = -16(t²– 2*9t/2 + 81/4) +160 + 324 = -16(t-9/2)² + 484. Thus the trajectory of the ball is a parabola opening downwards, with vertex at (9/2, 484). Since the vertex is the highest point a parabola opening downwards, the ball takes 9/2 = 4.5 seconds in reaching the highest point.