Height and fall of a ball

Question

A person standing close to the edge on top of a 48-foot building throws a ball vertically upward. The quadratic function ℎ(t)=-16t^2+52t+48 models the ball's height about the ground, ℎ(t), in feet, t seconds after it was thrown.

a) What is the maximum height of the ball?

b) How many seconds does it take until the ball hits the ground?

Answer 1

a). Maximum height of the ball:

1. Maximize the height h(t) = -16t² + 52t + 48
2. Find 1st derivative: h'(t) = -32t + 52
3. Find critical value when h'(t) = 0: Set -32t + 52 = 0 <=> t = 52/32 = 13/8 = 1.625
4. For all t < 1.625: h'(t) = -32t + 52 > 0
5. For all t > 1.625: h'(t) = -32t + 52 < 0
6. Since h'(t) > 0 for all t < 1.625 and h'(t) < 0 for all t > 1.625, then h(1.625) = -16(1.625)² + 52(1.625) + 48 = 90.25 (feet) is the absolute maximum value of the height.

b) For the ball to hit the ground, the height of the ball is 0. Meaning solve for t when h(t) = 0

1. Set h(t) = -16t² + 52t + 48 = 0
2. This is a quadratic equation. Use quadratic formula to solve it. Thus, x = (-b ±√(b²-4ac))/2a, where a = -16, b = 52, c = 48 (the coefficients of t², t¹, t⁰)
3. Thus, t = -3/4 or t = 4
4. Note, t ≥ 0 (the number of seconds) => Only accept t = 4
5. Thus, t = 4 (seconds)

Answer 2

a) To find the maximum, we want to find the height of the ball when the velocity is 0. This is because at the maximum, the velocity will change from positive (ball moving up) to negative (ball moving down) and will momentarily have a velocity of 0 at the maximum height. We have the function for height, h(t), which can be differentiated to find velocity. v(t)=h(t)/dt=-32t+52. Since we want to find where velocity is 0, set v(t) equal to 0. this gives us 0=-32t+52. Solving for t gives us t=1.625. This is the time at which the ball is at its maximum height. Plug it back into h(t) and this gives us the maximum height.

b)If we know the ball is thrown at t=0, we need to find the time at which it hits the ground to know how long it has been in the air. When the ball hits the ground, its height is 0. h(t)=0=-16t²+52t+48. Solve for t to find what time it hits the ground.