Question

A projectile that was launched straight up from the ground with an initial velocity of 48 ft/s returned to the ground after 3 s. The height of the projectile t seconds after launch is modeled by the function f(t)=−16t2+48t . What is the maximum height of the projectile, in feet?

Answer 1

The time to reach maximum height is given by the formula t=-b/2a, where b is the linear term's coefficient and a is the quadratic term's coefficient. Plugging in the values, we find t=-48/-32=1.5 seconds. We can plug this back into the quadratic equation to get h=f(1.5)=36 feet.

Answer 2

The maximum height attended by the projectile is 36 feet initial velocity of the projectile, $u=48ft/sec$; the total time of the projectile is $t=3sec$ and the function of the height of the project with respect to time, $f(t)=-16t^2+48t$.

Maximum value of the function

The maximum value that can be achieved by a function in the specified domain is known as the maxima of the function.

To evaluate the maximum or the minimum value of the function, we need to first evaluate the critical point of the function.

How to evaluate the maximum and minimum values of the functions?

The given function for the height of the project with respect to time is-$f(t)=-16t^2+48t$

Differentiate the function with respect to time-

$\dfrac{d(f(t))}{dt}=dfrac{d}{dt}(-16t^2+48t)\\=-32t+48$

Evaluate the differentiated function to 0, to determine the critaical value as-

$-32t+48=0\\t=\dfrac{48}{32}\\t=1.5\text{sec}$

Differentiate the first derivative of the function again,

$\dfrac{d^2(f(t))}{dt^2}=\dfrac{d}{dt}(-32t+48)\\=-32$

As the second derivative of the function is a negative value so, the critical value is the point of maxima.

Substitute the value of time, $t=1.5\text{sec}$ in the function of the height as-

$f(t=1.5)=-16(1.5)^2+48(1.5)\\=-36+72\\=36\text{ft}$

Thus, the maximum height attended by the projectile is $36\text{ft}$.

Learn more about maximum and minimum values of the functions here- https://brainly.com/question/14996337

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