Question

April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The pathway of the
arrow can be represented by the equation h = -16t2 + 80t + 25, where h is the height and t is the time in
seconds. What is the maximum height of the arrow?
90 feet
140 feet
125 feet
80 feet

Answer 1

Answer:

The maximum height of the arrow is 125 feet.

Step-by-step explanation:

The pathway of the arrow can be represented by the equation,

$h = -16t^2 +80t + 25$ .....(1)

Where h is height in in feet and t is time in seconds.

It is required to find the maximum height of the arrow. For maximum height, $\dfrac{dh}{dt}=0$.

So,

$\dfrac{d(-16t^2 +80t + 25)}{dt}=0\\\\-32t+80=0\\\\t=\dfrac{80}{32}\\\\t=2.5\ s$

Put t = 2.5 s in equation (1). So,

$h = -16(2.5)^2 +80(2.5) + 25\\\\h=125\ \text{feet}$

So, the maximum height of the arrow is 125 feet.