- 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.
Figure 1
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.