Question

If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h after t seconds is given by the equation h equals negative 16 t squared plus 112 t plus 128. After how many seconds does the object hit the​ ground? Round to the nearest tenth of a second.

Answer 1

Explanation:

The equation of motion of an object is given by :

$h(t)=-16t^2+112t+128$

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

$-16t^2+112t+128=0$

$-t^2+7t+8=0$

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.