Question

A projectile is launched from a hole in the ground one foot deep. It’s height follows the equation h= -16t^2 +8t-1. factoring by perfect squares to find the time when the projectile lands back on the ground.(Hint:Landing on the ground means projectile height is zero.)

Answer 1

Given:

Height of projectile from earth surface is -16 t² + 8 t - 1.

To Find:

Find the time when the projectile lands back on the ground.

Solution:

Height of projectile from earth surface is given:

$\mathbf{ -16t^{2}+ 8t -1=H}$

We have to find out time when it return to ground means H become zero:

$\mathbf{ -16t^{2}+ 8t -1=0}$

On  divide by -16 on both side:

$\mathbf{t^{2}- \frac{1}{2}t +\frac{1}{16}=0}$

This can be written as:

$\mathbf{t^{2}- 2\times \frac{1}{4}\times t +\left ( \frac{1}{4} \right )^{2}=0}$     ...1)

By identity formula:

$\mathbf{(A-B)^{2}=A^{2}-2\times A\times B+B^{2}}$      ...2)

On comparing equation 1) and equation 2), we can write equation 1) in term of;

$\mathbf{\left ( t-\frac{1}{4} \right )^{2}=0}$

Means:

$\mathbf{ t-\frac{1}{4} =0}$

So:

$\mathbf{ t=\frac{1}{4}= 0.25 \ hours }$

Means the projectile lands back on the ground after 0.25 hours after launch.