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

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

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.