Question

An object is fired straight up with a velocity of 32 ft./s. The height (h) reached by the object at (t) seconds is given by h(t) = -8t + 32t. When does the object hit the ground?

A) 8 seconds
B) 4 seconds
C) 6 seconds
D) 2 seconds

Answer 1

The object hits the ground 8 seconds after being fired straight up with a velocity of 32 ft./s

Step-by-step explanation:

To determine when the object hits the ground, we need to find the value of t when h(t) is equal to 0. Substituting h(t) = 0 into the equation -8t + 32t, we get:

0 = -8t + 32t

0 = 24t

t = 0

Therefore, the object hits the ground at t = 0 seconds. However, we know that the initial velocity is 32 ft./s, which means the object is already in motion. We can find the time it takes for the object to reach its maximum height by finding the time when its velocity is 0. The equation for velocity is v(t) = -8 + 32, which gives:

0 = -8t + 32

8t = 32

t = 4

Therefore, the object reaches its maximum height at t = 4 seconds. Since the object hits the ground at t = 0 seconds and reaches its maximum height at t = 4 seconds, we can conclude that the object hits the ground again at t = 8 seconds. Therefore, the correct answer is A) 8 seconds.