Question

question a tennis ball is dropped from a certain height. its height in feet is given by h(t)= - 16t² + 196 where t represents the time in seconds after launch. how long is the ball in the air?

Explanation:

Step1: Set height equal to 0

We want to find when the ball hits the ground, so we set $h(t)=0$. Thus, $- 16t^{2}+196 = 0$.

Step2: Rearrange the equation

Add $16t^{2}$ to both sides: $16t^{2}=196$.

Step3: Solve for $t^{2}$

Divide both sides by 16: $t^{2}=\frac{196}{16}=\frac{49}{4}$.

Step4: Solve for $t$

Take the square - root of both sides. Since $t\geq0$ (time cannot be negative in this context), $t=\sqrt{\frac{49}{4}}=\frac{7}{2}=3.5$.

Answer:

$3.5$ seconds