Question

An object is thrown upward from the top of an 80 ​-foot building with an initial velocity of 64 feet per second. The height h of the object after t seconds is given by the quadratic equation H= -16T^2+64T +80. When will the object hit the​ ground?

Answer 1

Answer:

The object will hit the ground 5 seconds after it is thrown.

Step-by-step explanation:

Given equation:

$h(t)=-16t^2+64t+80$

where:

• h is the height of the object from the ground (in feet).
• t is time (in seconds).

The object will hit the ground when the height of the object is zero.

Therefore, to find the time when the object hits to ground, set the equation to zero and solve for t.

$\begin{aligned}\implies -16t^2+64t+80&=0\\-16(t^2-4t-5)&=0\\t^2-4t-5&=0\\t^2-5t+t-5&=0\\t(t-5)+1(t-5)&=0\\(t+1)(t-5)&=0\\\\t+1&=0\implies t=-1\\t-5&=0\implies t=5\end{aligned}$

As t ≥ 0, t = 5 only.

Therefore, the object will hit the ground 5 seconds after it is thrown.