Question

An object is thrown up word from the top of 160 foot building with an initial velocity of 144 ft./s. The height H of the object after T seconds is given by the quadratic equation h= -16t^2 + 144t + 160. When will the object hit the ground?

Answer 1

Answer:

After 10 seconds

Step-by-step explanation:

In this problem, the height of the object after t seconds is described by the function

$h(t)=-16t^2+144t+160$

where

160 ft is the initial height of the object at t = 0

+144 ft/s is the initial velocity

$-32 ft/s^2$ is the acceleration due to gravity (downward)

Here we want to find the time at which the object hits the ground, so the time t at which

$h(t)=0$

Therefore we can write

$-16t^2+144t+160 =0$

Simplifying (dividing each term by 16), we get

$-t^2+9t+10=0$

This is a second-order equation in the form

$ax^2+bx+c=0$

Which has solutions given by the formula

$t_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (2)

Here we have:

a = -1

b = 9

c = 10

Substituting into (2) we find the solutions:

$t_{1,2}=\frac{-9\pm \sqrt{9^2-4(-1)(10)}}{2(-1)}=\frac{-9 \pm 11}{-2}$

Which gives:

$t_1=10 s\\t_2 =-1 s$

Since time cannot be negative, the only solution is

t = 10 seconds