**The distance $s$ of the ball from the ground after $t$ sec is $s(t)= 96t-16t^2$.****At what time $t$ will the ball strike the ground?****For what time $t$ is the ball more than $128$ feet above the ground?**

The aim of this question is to find the **time $t$** in which the **ball** will hit the **ground** and the time $t$ after which it will be **$128$ feet** above the **ground.**

This question is based on the concept of **Torricelli’s Equation** **for accelerated motion **which is represented as follows:

\[V^2 = V_{\circ}^2 \times 2a\Delta S \]

Here,

**$V$= Final velocity**

**$V_{\circ}$= Initial velocity**

**$a$ = acceleration,** which is **gravitational acceleration** in this case ($a =g= 9.8 \dfrac {m}{s^2}$ or $32\dfrac{ft} {s^2}$)

**$\Delta S$ = distance traveled by the ball**

## Expert Answer

$(a)$ To find the **time** $t$ for which the ball will hit the ground, we will put the **function** of **distance** equal to zero because the **final distance** from the ground will be **zero,** so it will be written as:

\[s(t)= 96t-16t^2 = 0\]

\[96t-16t^2 = 0\]

\[t \left( 96-16t \right ) = 0\]

We get **$2$** equations:

\[t =0\] and \[ 96-16t=0\]

\[ -16t=-96\]

\[ t=\frac{-96}{-16}\]

\[t= 6\]

So we get **$t=0 sec$** and **$t=6 sec$. **Here, **$t=0$** when the **ball** is at **rest **and **$t=6 sec$** is when the ball comes back to the ground after being **thrown upwards.**

$(b)$ To find the **time** $t$ for which it will be $128$ feet above the ground, we will put the function equal to $128$, which is the given distance.

\[s(t)= 96t-16t^2 \]

\[128= 96t-16t^2 \]

\[0= 96t-16t^2 -128 \]

\[16t^2 -96t+128 =0 \]

Taking $16$ common

\[16\left (t^2 -6t+8 \right) =0 \]

\[t^2 -6t+8 =0\]

Making factors, we get:

\[t^2 -4t-2t+8 =0\]

\[t \left( t -4\right)-2\left( t -4\right) =0\]

\[ \left( t -4\right)\times \left( t -2\right) =0\]

We get:

\[t=4 sec \] and \[t =2 sec\]

Thus, the** time** $t$ for which the ball will be **$128$ feet** above the ground is between the time **$t= 4sec$** and **$t=2 sec$.**

## Numerical Result

The **time** $t$ for which the ball will **hit** the **ground** is calculated as:

\[t = 6 sec\]

Thus, the **time $t$** for which the ball will be **$128$** feet above the ground is between the time **$t= 4sec** $ and **$t=2 sec$.**

## Example

A **rock** is thrown **vertically upward** with an initial **velocity** of **$80$ feet** per **second.** The **distance $s$** of the rock from the ground after **$t$ sec** is **$s(t)= 80t-16t^2$.** At what time **$t$** will the rock **strike** the **ground?**

Given the **function** of **distance,** we will put it equal to zero as:

\[s(t)= 80t-16t^2 = 0\]

\[80t-16t^2 = 0\]

\[t \left( 80-16t \right ) = 0\]

We get **$2$** equations:

\[t =0\] and \[ 80-16t=0\]

\[-16t=-80\]

\[ t=\frac{-80}{-16}\]

\[t= 5\]

so we get $t=0 sec$ and $t=5 sec$.

Here, **$t=0$** is when the rock is at rest initially,

and **$t=5 sec$** is when the **rock** comes back to the **ground** after it is** thrown upwards.**