Question

A golf ball is hit with an initial vertical velocity of 80 FPS. H= -16t^2+ 80t. How high is the ball after 2 seconds

Answer 1

Answer:

The height of the ball after 2 seconds = 96 feet.

Step-by-step explanation:

Given:

The height of a golf ball hit with an initial velocity of 80 FPS is given by:

$H=-16t^2+80t$

where $H$ represents height of the ball in feet and $t$ represents time in seconds.

To find the height of the ball after 2 seconds.

Solution:

In order to find the height of the ball after 2 seconds, we plugin t=2 in the given function of height.

We have:

$H(t)=-16t^2+80t$

At $t=2$ seconds

$H(2)=-16(2)^2+80(2)$

$H(2)=-16(4)+160$

$H(2)=-64+160$

$H(2)=96$

Thus, height of the ball after 2 seconds = 96 feet.