Question

A ball is launched out of a cannon that is 24 feet off the ground. The height of the ball after "t" seconds is represented by

Answer 1

Given:

The height of ball is represented by the below function:

$h(t)=-16t^2+20t+24$

To find:

The number of seconds it will take to reach the ground.

Solution:

We have,

$h(t)=-16t^2+20t+24$

At ground level, the height of ball is 0, i.e., $h(t)=0$.

$-16t^2+20t+24=0$

Taking out greatest common factor.

$-4(4t^2-5t-6)=0$

$4t^2-5t-6=0$

Splitting the middle term, we get

$4t^2-8t+3t-6=0$

$4t(t-2)+3(t-2)=0$

$(t-2)(4t+3)=0$

Using zero product property, we get

$(t-2)=0$ and $(4t+3)=0$

$t=2$ and $t=-\dfrac{3}{4}$

Time cannot be negative, so $t=2$.

Therefore, the ball will reach the ground after 2 seconds.