Question

An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t2, 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? When will the object hit the ground?

Answer 1

Answer:

• The object will be 144 feet in the air at t= 3 seconds

• At t= 6 seconds, the object will hit the ground.

Step-by-step explanation:

We have been the model $h(t) = 96t -16t^2$

Here.  t is the time since launch and h(t) is the height in feet of the projectile after time t in seconds.

For the height of the object to be 144 feet, we have

$h(t)=144\\\\96t -16t^2=144\\\\16t^2-96t+144=0\\\\\text{Using the quadratic formula, we get}\\\\t_{1,\:2}=\frac{-\left(-96\right)\pm \sqrt{\left(-96\right)^2-4\cdot \:16\cdot \:144}}{2\cdot \:16}\\\\t_{1,\:2}=\frac{-\left(-96\right)\pm \sqrt{0}}{2\cdot \:16}\\\\t=3$

Therefore, after 3 seconds the object will be 144 feet in the air.

Now, when the object hits the ground then the height should be zero.

$96t -16t^2=0\\16t(6-t)=0\\t=0,6$

Hence, at t= 6 seconds, the object will hit the ground.

Answer 2

Answer to Q1 is it will be 144 feet high in the air : after 3.0 seconds
Answer to Q2 is it will hit the groundat  about: after 6.0 seconds