Question

A ball is thrown vertically in the air with a velocity of 95ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 120ft. Round your answer(s) to the nearest tenth of a second.

Answer 1

The Answer Was "To the nearest tenth of a second, the ball was in the air for 2.4s"

Explanation:

Answer 2

`t_1 = 1.8\ s\\\\t_2 = 4.1\ s`

Explanation:

If the equation
`h = -16t^2 + v_0t` represents the position of the ball as a function of time then, to find in which second the ball reaches 120 feet must substitute
`h = 120` in the equation of the height and solve for t.

`120 = -16t ^ 2 + v_0t`

If the initial velocity is 95 feet/s then
`v_0 = 95`

Then:

`120 = -16t ^ 2 + 95t\\\\-16t ^ 2 + 95t -120 = 0`

Use the quadratic formula

`t_1 = (-b+√(b^2 -4ac))/(2a)\\\\t_2 = (-b-√(b^2 -4ac))/(2a)`

Where, for this problem:

`a = -16\\b = 95\\c = -120`

So

`t_1 =(-95+√((95)^2 -4(-16)(-120)))/(2(-16)) = 1.8\ s\\\\t_2=(-95+√((95)^2 -4(-16)(-120)))/(2(-16)) = 4.1\ s`

This result means that the ball reaches 120 feet for the first time at 1.8 seconds, then begins to descend and on its descent again reaches 120 feet at t = 4.1 seconds.