A ball is thrown downward with a speed of 8 ft/s from the top of a 68 ft tall building. After t seconds, its height above the ground is given by s(t) = −16t^2 − 8t + 68.

Question

A ball is thrown downward with a speed of 8 ft/s from the top of a 68 ft tall building. After t seconds, its height above the ground is given by s(t) = −16t² − 8t + 68.

(a)Determine how long (in s) it takes for the ball to hit the ground.

(b)Determine the velocity (in ft/s) of the ball when it hits the ground.

Answer 1

(a)

We want to figure out the time when the ball hits the ground — in other words, we want to know what t is when s(t) = 0. So, we want to solve

-16t² - 8t + 68 = 0

This equation is not easily factorable, so let's use the quadratic equation to solve it. The quadratic equation tells us that

t = (8 ± sqrt(8² - 4*(-16)*68)) / (2*(-16))

This gives us two solutions for t:

t ≈ 1.827 s and t ≈ -2.327 s

These both solve the equation s(t) = 0, but only one of them makes sense. In particular, because time can't be negative, we have to take the positive solution. So the answer is t ≈ 1.827 s.

(b)

Now, we want to know what the velocity of the ball is when it hits the ground. In other words, if v(t) is the velocity of the ball at time t, we want to know v(1.827).

We don't have an equation for velocity given to us, but we know that the velocity curve is the derivative of the position curve:

v(t) = s'(t) = -32t - 8

Now, we can compute:

v(1.827) = -32*1.827 - 8 ≈ -66.453 m/s.