Question

An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t squared plus 120 t describes the height of the debris above the​ ground, h, in​ feet, t seconds after the explosion. When will the debris be 56 feet above the​ ground?

Answer 1

Answer:

The debris will be at a height of 56 ft when time is 0.5 s and 7 s.

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$. Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$

Therefore, the debris will reach a height of 56 ft twice.

When time $t=0.5\ s$ during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at $t=7\ s$, the height is 56 ft.