Chapter 3.5, Problem 33AYU

Physics A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance $s$ (in feet) of the ball from the ground after $t$ seconds is $s\left(t\right)=80t-16t^2$.

(a) At what time $t$ will the ball strike the ground?

(b) For what time $t$ is the ball more than 96 feet above the ground?

To determine

To find:

a. The time $t$ at which the ball strike the ground.

b. The time $t$ at which the ball is more than 96 feet above the ground.

Answer to Problem 33AYU

Solution:

a. After 5 seconds the ball will strike the ground.

b. For the time $2<t<3$, the ball will be more than 96 feet above the ground level.

Explanation of Solution

Given:

The distance of the ball from the ground after $t$ seconds is given by:

$s\left(t\right)=80t-16t^2$

Calculation:

a. $s\left(t\right)=80t-16t^2$

To find the time $t$ at which the ball will hit the ground is found by solving the equation:

$s\left(t\right)=80t-16t^2=\text{ 0}$

$16t\left(5-t\right)=0$

$t=5$

Thus after 5 seconds the ball will strike the ground.

b. The ball is more than 96 feet above the ground.

$s\left(t\right)=80t-16t^2>96$

$16t^2-80t+96<0$

$16\left(t^2-5t+6\right)<0$

$t^2-5t+6<0$

$t^2-2t-3t+6<0$

$t\left(t-2\right)-3\left(t-2\right)<0$

$\left(t-2\right)\left(t-3\right)<0$

$2<t<3$

Thus for the time $t$ $\left(2<t<3\right)$ the ball will be more than 96 feet above the ground.