Question

When a baseball is hit by a batter, the height of the ball, h(t), at time t, t=0 is determined by the equation h(t)=-16t^2+64t^2+4. If t is in seconds, for which interval of time is the height of the ball greater than or equal to 52 feet?
A: 1 B: t<=1
C: 0 D: 1<=t<=3

Answer 1

Let

t--------> the time in seconds

h--------> the height of the ball in feet

we know that

$h(t)=-16t^2+64t^2+4$

Using a graphing tool

see the attached figure

The height of the ball is greater than or equal to $52$ feet for

$t\geq 1\ sec$

therefore

the answer is the option D

$1\ sec \leq t \leq 3\ sec$

Answer 2

The correct option is $\boxed{\bf\ option D}$ i.e., $\boxed{1\leq t\leq 3}$.

Further explanation:

Given:

The height of the ball is determined by the equation $h(t)=-16t^{2}+64t^{2}+4$.

Calculation:

Consider $t$ as the interval of time and $h$ as the height of the time.

The equation is given as follows:

$\boxed{h(t)=-16t^{2}+64t^{2}+4}$

Further solve the above equation as follows:

$\begin{aligned}h(t)&=-16t^{2}+64t^{2}+4\\&=48t^{2}+4\end{aligned}$

We have to check the interval in which height of the ball is greater than or equal to $52\text{ feet}$.

Check the option A by substituting $t=1$ in the equation $h(t)=-16t^{2}+64t^{2}+4$ as follows:

$\begin{aligned}h(1)&=48\cdot (1)^{2}+4\\&=18+4\\&=52\end{aligned}$

The height of the ball is $52\text{ feet}$ at $t=1$.

The option A is not full filling the condition completely as the height is not greater than $52\text{ feet}$.

This implies that option A is incorrect.

Now, move to the option B that is $t\leq 1$.

Check the second option by substituting $t=0$ in the equation $h(t)=48t^{2}+4$ as follows:

$\begin{aligned}h(0)&=48\cdot (0)^{2}+4\\&=0+4\\&=4\end{aligned}$  

The height of the ball is $4\text{ feet}$ at time $t=0$.

The inequality for time $t\leq 1$ is not satisfying the condition.

Therefore, option B is incorrect.

Now check for the option C that is $t=0$ but we have already discarded this option in option B.

So, there is no need to check the option C.

Therefore, check the last option D that is the interval of time $1\leq t\leq 3$.

The height of the ball is $52\text{ feet}$ at time $t=1$ as we already calculated for the option A.

Now substitute $t=3$ in the given equation  to check the interval $1\leq t\leq 3$.

$\begin{aligned}h(3)&=48\cdot (3)^{2}+4\\&=(48\cdot 9)+4\\&=436\end{aligned}$  

The height of the ball is $436\text{ feet}$ at time $t=3$ satisfies the condition completely as the height $436\text{ feet}$ is greater than $52\text{ feet}$.

Therefore, the boundary points of the interval $1\leq t\leq 3$ satisfy the condition.

Now check another point which lies in the interval of time $1\leq t\leq 3$.

So, substitute $t=2$ in the equation $h(t)=48t^{2}+4$.

$\begin{aligned}h(2)&=48\cdot (2)^{2}+4\\&=(48\cdot 4)+4\\&=196\end{aligned}$  

The height of the ball is $196\text{ feet}$ at time $t=2$ satisfies the condition completely as the height $196\text{ feet}$ is greater than $52\text{ feet}$.

Therefore, the height of the ball is greater than or equal to $52\text{ feet}$ in the interval of time $1\leq t\leq 3$.

Thus, the correct option is $\boxed{\bf option D}$.

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Quadratic equations

Keywords: Polynomial, quadratic equation, substitution, baseball, interval, height, greater than, equation, time, seconds, simplification, solution, inequality.