Question

A policeman shoots a bullet with an initial vertical velocity of 128 feet per second from a building that is 68 feet away from the ground. Use the vertical motion model, h = -16t^2vt+s where v is the initial velocity in feet per second and s is the height in feet, to calculate how long the bullet will be in the air for. Round your answer to the nearest tenth.

Answer 1

h is the height at time t

t is the time

v is initial vertical velocity

s is initial height

s = 68ft, v = 128ft/s

Therefore

h = -16t^2 + 128t + 68

The time the ball is in the air is the time it took the bullet to hit the ground (h = 0)

To find the time taken for the bullet to be in the air,

we set h = 0.

`\Rightarrow-16t^2+128t+68=0`

Dividing through by -16, we have;

`\begin{gathered} t^2-8t-(17)/(4)=0 \\ By\text{ completing the square method, we have} \\ (t-4)^2-(-4)^2-(17)/(4)=0 \\ (t-4)^2-(81)/(4)=0 \\ \Rightarrow t-4=\pm\sqrt[]{(81)/(4)} \\ \Rightarrow t=4\pm(9)/(2)=4-4.5\text{ or 4+4.5} \\ t=-0.5\text{ or 8.5} \end{gathered}`

Since t cannot be negative, then the only possibility is

t = 8.5s

Hence the bullet was in the air for 8.5s