Question

A ball is thrown upward with a speed of 44ft/sec from the top of a 64 -foot-tall building. After t seconds, its height above the ground is given by (8pts) s(t)=−16t² +44t+64 feet at t seconds

a) Find the velocity function.
b) Determine when the ball's velocity is zero. (write as a fraction)
c) Determine the maximum height reached by the ball. (write as a fraction)
d) Determine how long it takes for the ball to reach the ground. (round to 2 places)

Answer 1

The velocity function of the ball is v(t) = -32t + 44. The ball's velocity is zero after 11/8 seconds. The maximum height reached by the ball is 121/8 feet. It takes approximately 3.94 seconds for the ball to reach the ground.

Step-by-step explanation:

The velocity function of the ball can be found by taking the derivative of the height function with respect to time. The derivative of -16t² + 44t + 64 is -32t + 44. So, the velocity function is v(t) = -32t + 44 feet per second.

To determine when the ball's velocity is zero, we set v(t) = 0 and solve for t. -32t + 44 = 0, t = 11/8. So, the ball's velocity is zero after 11/8 seconds.

The maximum height reached by the ball can be found by determining the vertex of the height function. The vertex occurs at -b/2a. In this case, a = -16 and b = 44. So, the maximum height is -(-44)/(2*-16) = 121/8 feet.

To determine how long it takes for the ball to reach the ground, we set s(t) = 0 and solve for t. -16t² + 44t + 64 = 0. Using the quadratic formula, we find that t is approximately 3.94 seconds.