Question

An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t^2, where t is the time since launch and h(t) is the height in feet of the projectile after time t in seconds. When will the object be 144 feet in the air?

A) After 1.5 seconds
B) After 3.0 seconds
C) After 6.0 seconds
D) After 2.5 seconds

When will the object hit the ground?

A) After 2.5 seconds
B) After 5.0 seconds
C) After 6.0 seconds

Answer 1

The object will be 144 feet in the air after 3.5 seconds and will hit the ground after 6 seconds.

Step-by-step explanation:

The object will be 144 feet in the air when h(t) is equal to 144. So we can set up the equation 96t - 16t^2 = 144 to solve for t. Rearranging the equation, we get 16t^2 - 96t + 144 = 0.

Using the quadratic formula, we find that t = 3 seconds and t = 3.5 seconds. However, we can disregard t = 3 seconds since it is outside the given time range. Therefore, the object will be 144 feet in the air after 3.5 seconds, which corresponds to option D.

The object will hit the ground when h(t) = 0. So we can set up the equation 96t - 16t^2 = 0. Factoring out t, we get t(96 - 16t) = 0. This gives us two solutions: t = 0 and t = 6 seconds. However, t = 0 corresponds to the time at launch and we can disregard it. Therefore, the object will hit the ground after 6 seconds, which corresponds to option C.