Question

A ball is thrown vertically upward with an initial velocity of 64 feet per second. The distance s​ (in feet) of the ball from the ground after t seconds is s=64t-16t^2 (a) At what time t will the ball strike the​ ground? ​(b) For what time t is the ball more than 48 feet above the​ ground?

Answer 1

Given:

The distance s​ (in feet) of the ball from the ground after t seconds is

$s=64t-16t^2$

To find:

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

(b) For what time t is the ball more than 48 feet above the​ ground?

Solution:

(a)

We have,

$s=64t-16t^2$

Substitute s=0, to find the time t when the ball strike the​ ground.

$0=64t-16t^2$

$0=16t(4-t)$

Using zero product property, we get

$16t=0\Rightarrow t=0$

$4-t=0\Rightarrow t=4$

Therefore, the ball strike the​ ground in initial condition (t = 0) and after 4 seconds (t = 4).

(b)

Now, s > 48, to find the time t when the ball will be more than 48 feet above the​ ground.

$64t-16t^2\geq 48$

$0> 16t^2-64t+48$

Divide both sides by 16.

$0> t^2-4t+3$

$0> t^2-t-3t+3$

$0> t(t-1)-3(t-1)$

$0>(t-1)(t-3)$

Related equation is $(t-1)(t-3)=0$. Zeroes are t=1,3. These two number divide the number line is three parts. (-∞,1),(1,3),(3,∞)

$0>(t-1)(t-3)$ inequality is true for only (1,3).

It is only possible when t lies in the interval (1,3).

Therefore, the ball will be more than 48 feet above the​ ground between 1 and 3 seconds.