Question

An arrow is shot vertically upward at a rate of 240 feet per second from ground level. Use the projectile formula to determine when the arrow hit the ground.

Answer 1

After 14.92 seconds, the arrow hits the ground.

Explanation:

Given:

Initial velocity of the arrow is,
`u=240\ ft/s or 73.152\ m/s`

The arrow goes up and come back to the same starting position. So, initial position is same as that of final position.

Therefore, displacement of the arrow is 0 m.

The displacement of the arrow is given as:

`s=ut-(1)/(2)gt^(2)` (equation 1)

Plug all value in equation 1 and solve for
`t`

`0=73.152*ts t-(1)/(2) *9.8 * t^(2)`

`0=73.152*ts t-4.9 * t^(2)`

`0=t(73.152-4.9 * t)`

`73.152-4.9* t=0` where
`t\\eq 0`

`4.9* t=73.152`

`t=(73.152)/(4.9)= 14.92\ s`

∴ After 14.92 seconds, the arrow hits the ground.