Question

A ball is thrown upward at a velocity of 60 meters per second from a height of 80 feet above the ground. The height h (in meters) of the ball at time t (in seconds) after it is thrown can be found by h(t) = -20t^2 + 60t + 80. Find the time when the ball is again 80 feet above the ground.

A. 0.5 seconds
B. 1.5 seconds
C. 2.5 seconds
D. 3.5 seconds

Answer 1

The time when the ball is again 80 feet above the ground is 3 seconds.

Step-by-step explanation:

To find the time when the ball is again 80 feet above the ground, we need to set the equation for height equal to 80 and solve for t. So, we have:
h(t) = -20t^2 + 60t + 80
Setting h(t) = 80:
-20t^2 + 60t + 80 = 80
Subtracting 80 from both sides:
-20t^2 + 60t = 0
Factoring out t:
t(-20t + 60) = 0
Setting each factor equal to 0:
t = 0
-20t + 60 = 0
Solving for t, we find t = 3 seconds.
Therefore, the time when the ball is again 80 feet above the ground is 3 seconds.