Question

A ball is thrown into the air with an upward velocity of 26 feet per second from a building that is 12 feet high. The equation that gives the height of the ball above the ground at time t is

h = 12 + 26t - 16+2
a. At what time(s) will the ball be 22 feet above the ground?
b. When will the ball hit the ground?

Answer 1

Answer: A. The ball will be 22 feet above the ground at times t = 5/8 seconds and t = 1 second

b. The ball will hit the ground at time t = 2 seconds.

Explanation:

A.h = 12 + 26t - 16t^2

22 = 12 + 26t - 16t^2

16t^2 - 26t + 10 = 0

8t^2 - 13t + 5 = 0

(8t - 5)(t - 1) = 0

So the solutions are t = 5/8 and t = 1. Therefore, the ball will be 22 feet above the ground at times t = 5/8 seconds and t = 1 second.

b. 0 = 12 + 26t - 16t^2

16t^2 - 26t - 12 = 0

8t^2 - 13t - 6 = 0

(8t + 3)(t - 2) = 0

So the solutions are t = -3/8 and t = 2. We can ignore the negative solution, since time cannot be negative in this context. Therefore, the ball will hit the ground at time t = 2 seconds.