Question

A ball is thrown vertically upward with an initial velocity of 80 feet per second. The

distances (in feet) of the ball from the ground after t seconds is s = 80t - 16t².
(a) At what time t will the ball strike the ground?
(b) For what time t is the ball more than 84 feet above the ground?

Answer 1

The ball strikes the ground at t = 5 seconds. The ball is more than 84 feet above the ground for t values between 3 and 7 seconds.

Step-by-step explanation:

(a) To find the time at which the ball strikes the ground, we need to set the distance from the ground equal to zero and solve for t.

Using the equation s = 80t - 16t², we set s = 0 and solve for t:

0 = 80t - 16t²

16t² - 80t = 0

16t(t - 5) = 0

t = 0 or t = 5

Since time can't be negative, the ball strikes the ground at t = 5 seconds.

(b) To find the time at which the ball is more than 84 feet above the ground, we need to set the distance from the ground greater than 84 and solve for t.

Using the equation s = 80t - 16t², we set s > 84 and solve for t:

80t - 16t² > 84

16t² - 80t + 84 < 0

4(t - 3)(t - 7) < 0

3 < t < 7

Therefore, the ball is more than 84 feet above the ground for t values between 3 and 7 seconds.