Question

A man throws a ball into the air with a velocity of $96feet/sec$.Use the formula $h=-16t^2+v_{0}t$to determine when the height of the ball will be 48 feet . Round to the nearest tenth of a second.

Short answer

At 0.6 seconds as well as 5.4 seconds the height of the ball will be 48 feet .

Step-by-step solution

Given information 

The given statement is a man throws a ball into the air with a velocity of 96 feet per second.

Solve the equation 

$$\begin{gathered} h=-16t^2+v_{0}t \\ \mathrm{height}\mathrm{is}48\mathrm{feet} \\ \mathrm{v}=96\mathrm{ft}/\sec \\ \mathrm{Substitute}\mathrm{the}\mathrm{values} \\ 48=-16\left({\mathrm{t}}^2\right)+\left(96\right)\mathrm{t} \\ \mathrm{Take}\mathrm{out}16\mathrm{as}\mathrm{a}\mathrm{common} \\ 3=-{\mathrm{t}}^2+6\mathrm{t} \\ {\mathrm{t}}^2-6\mathrm{t}+3=0 \end{gathered}$$

Quadratic equation 

$$\begin{gathered} t^2-6t+3=0 \\ a=1,b=-6andc=3 \\ t=\frac{6\pm \sqrt{36-4\left(3\right)}}{2\left(1\right)} \\ t=\frac{6\pm \sqrt{36-12}}{2} \\ t=\frac{6\pm \sqrt{24}}{2} \\ t=\frac{6\pm 4.8}{2} \\ t=0.6or5.4 \end{gathered}$$

At 0.6 seconds as well as 5.4 seconds the height of the ball will be 48 feet .