Question

Vikas and Vasu punt a ball into the air. The equation $h=-16{{t}^{2}}+60t$ represents the height of the ball in feet, $t$ seconds after it was punted for Vikas’s ball. Which of the following can be Akanksha’s ball height equation if her ball goes higher?
A. $h=-16\left( {{t}^{2}}-3t \right)$
B. $h=-8t\left( 2{{t}^{2}}-9 \right)$
C. $h=-4{{\left( 2{{t}^{2}}-5 \right)}^{2}}+48$
D. $h=-4{{\left( 2{{t}^{2}}-6 \right)}^{2}}+52$

Answer 1

Hint: In this problem they have mentioned the equation for the trajectory of the ball. Now we will differentiate the equation and equate it to zero to know at which time that the ball has maximum height. After finding the value of $t$ for the maximum height we will calculate the maximum height by substituting the value of $t$ in the given equation. Now we have the maximum height of the ball punt by Vikas. Now we will do the same procedure and find the maximum height of the ball for each given option. Now we will compare the values of four options with the maximum height of the ball punt by Vikas to get the result.

Complete step-by-step answer:
Given that, the trajectory of the ball punt by Vikas is
$h\left( t \right)=-16{{t}^{2}}+60t$
Differentiating the above equation with respect to $t$, then we will get
$\begin{align}
  & {{h}^{'}}\left( t \right)=\dfrac{d}{dt}\left( -16{{t}^{2}}+60t \right) \\
 & \Rightarrow {{h}^{'}}\left( t \right)=-16\dfrac{d}{dt}\left( {{t}^{2}} \right)+60\dfrac{d}{dt}\left( t \right) \\
\end{align}$
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, then we will get
$\begin{align}
  & {{h}^{'}}\left( t \right)=-16\left( 2t \right)+60\left( 1 \right) \\
 & \Rightarrow {{h}^{'}}\left( t \right)=-32t+60....\left( \text{i} \right) \\
\end{align}$
Again, differentiating the above equation with respect to $t$, then we will have
$\begin{align}
  & {{h}^{''}}\left( t \right)=\dfrac{d}{dt}\left( -32t+60 \right) \\
 & \Rightarrow {{h}^{''}}\left( t \right)=-32\dfrac{d}{dt}\left( t \right)+0 \\
 & \Rightarrow {{h}^{''}}\left( t \right)=-32<0 \\
\end{align}$
We got ${{h}^{''}}\left( t \right)<0$ so the function $h\left( t \right)$ will have maximum. To find the maximum value we need to find the value of $t$ and that can be calculated by equating ${{h}^{'}}\left( t \right)$ to Zero.
$\begin{align}
  & \Rightarrow {{h}^{'}}\left( t \right)=0 \\
 & \Rightarrow -32t+60=0 \\
 & \Rightarrow 32t=60 \\
 & \Rightarrow t=\dfrac{15}{8} \\
\end{align}$
$\therefore $ We have maximum value of $h\left( t \right)$ at $t=\dfrac{15}{8}$ and that maximum value/height can be
$\begin{align}
  & h\left( t \right)=h\left( \dfrac{15}{8} \right) \\
 & \Rightarrow h\left( \dfrac{15}{8} \right)=-16{{\left( \dfrac{15}{8} \right)}^{2}}+60\left( \dfrac{15}{8} \right) \\
 & \Rightarrow h\left( \dfrac{15}{8} \right)=56.25 \\
\end{align}$
$\therefore $ Maximum height of the ball punted by Vikas is $56.25$ feet.
Now the maximum height of the balls that follows the equations given in options are

A.
$h=-16\left( {{t}^{2}}-3t \right)$
Differentiating the above equation with respect to $t$, then we will get
$\begin{align}
  & {{h}^{'}}\left( t \right)=\dfrac{d}{dt}\left( -16{{t}^{2}}+48t \right) \\
 & \Rightarrow {{h}^{'}}\left( t \right)=-16\dfrac{d}{dt}\left( {{t}^{2}} \right)+48\dfrac{d}{dt}\left( t \right) \\
\end{align}$
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, then we will get
$\begin{align}
  & {{h}^{'}}\left( t \right)=-16\left( 2t \right)+48\left( 1 \right) \\
 & \Rightarrow {{h}^{'}}\left( t \right)=-32t+48....\left( \text{i} \right) \\
\end{align}$
Again, differentiating the above equation with respect to $t$, then we will have
$\begin{align}
  & {{h}^{''}}\left( t \right)=\dfrac{d}{dt}\left( -32t+48 \right) \\
 & \Rightarrow {{h}^{''}}\left( t \right)=-32\dfrac{d}{dt}\left( t \right)+0 \\
 & \Rightarrow {{h}^{''}}\left( t \right)=-32<0 \\
\end{align}$
We got ${{h}^{''}}\left( t \right)<0$ so the function $h\left( t \right)$ will have a maximum. To find the maximum value we need to find the value of $t$ and that can be calculated by equating ${{h}^{'}}\left( t \right)$ to Zero.
$\begin{align}
  & \Rightarrow {{h}^{'}}\left( t \right)=0 \\
 & \Rightarrow -32t+48=0 \\
 & \Rightarrow 32t=48 \\
 & \Rightarrow t=\dfrac{3}{2} \\
\end{align}$
$\therefore $ We have maximum value of $h\left( t \right)$ at $t=\dfrac{3}{2}$ and that maximum value/height can be
$\begin{align}
  & h\left( \dfrac{3}{2} \right)=-16\left( {{\left( \dfrac{3}{2} \right)}^{2}}-3\left( \dfrac{3}{2} \right) \right) \\
 & \Rightarrow h\left( \dfrac{3}{2} \right)=36 \\
\end{align}$
$\therefore $ Maximum height of the ball that follow the equation $h=-16\left( {{t}^{2}}-3t \right)$ is $36$ feet which is less than the maximum height of Vikas’s ball.

B.
$h=-8t\left( 2{{t}^{2}}-9 \right)$
Differentiating the above equation with respect to $t$, then we will get
$\begin{align}
  & {{h}^{'}}\left( t \right)=\dfrac{d}{dt}\left( -16{{t}^{3}}+72t \right) \\
 & \Rightarrow {{h}^{'}}\left( t \right)=-16\dfrac{d}{dt}\left( {{t}^{3}} \right)+72\dfrac{d}{dt}\left( t \right) \\
\end{align}$
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, then we will get
$\begin{align}
  & {{h}^{'}}\left( t \right)=-16\left( 3{{t}^{2}} \right)+72\left( 1 \right) \\
 & \Rightarrow {{h}^{'}}\left( t \right)=-48{{t}^{2}}+72...\left( \text{i} \right) \\
\end{align}$
Again, differentiating the above equation with respect to $t$, then we will have
$\begin{align}
  & {{h}^{''}}\left( t \right)=\dfrac{d}{dt}\left( -48{{t}^{2}}+72 \right) \\
 & \Rightarrow {{h}^{''}}\left( t \right)=-48\dfrac{d}{dt}\left( {{t}^{2}} \right)+0 \\
 & \Rightarrow {{h}^{''}}\left( t \right)=-96t<0 \\
\end{align}$
We got ${{h}^{''}}\left( t \right)<0$ so the function $h\left( t \right)$ will have maximum. To find the maximum value we need to find the value of $t$ and that can be calculated by equating ${{h}^{'}}\left( t \right)$ to Zero.
$\begin{align}
  & \Rightarrow {{h}^{'}}\left( t \right)=0 \\
 & \Rightarrow -48{{t}^{2}}+72=0 \\
 & \Rightarrow 32t=72 \\
 & \Rightarrow t=\sqrt{1.5} \\
\end{align}$
$\therefore $ We have maximum value of $h\left( t \right)$ at $t=\sqrt{1.5}$ and that maximum value/height can be
$\begin{align}
  & h\left( \sqrt{1.5} \right)=-8\left( \sqrt{1.5} \right)\left( 2{{\left( \sqrt{1.5} \right)}^{2}}-9 \right) \\
 & \Rightarrow h\left( \sqrt{1.5} \right)=58.787 \\
\end{align}$
$\therefore $ Maximum height of the ball that follow the equation $h=-8t\left( 2{{t}^{2}}-9 \right)$ is $58.787$ feet which is higher than the maximum height of Vikas’s ball.

So, the correct answer is “Option B”.

Note: Students may do mistake while calculating the maximum height for the equation in option-b since it has $t$ in multiplication with the ${{t}^{2}}$, student didn’t consider this and writes the equation as $h=-16{{t}^{2}}-9t$ if you do like this we don’t get the correct answer. Here we got the correct answer in option-b. If we don’t get here, we need to solve the equations in remaining options.