Question

A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second .

Answer 1

Step-by-step explanation:

Given that,

Velocity of the ball in the air is , v = 95 ft/s

We need to find the time when the ball is at a height of 120 feets. The equation of projectile is given by the following formula :

$h(t)=-16t^2+vt$

Here, h(t) = 120 feets

$-16t^2+95t-120=0$    

On solving above equation, we get the value of t as :

t = 1.823 s

t = 4.115 s

Therefore, at 1.823 seconds and 4.115 seconds the ball is at a height of 120 feets.

Answer 2

Answer:

On the way up, 1.8 seconds and on the way down 4.1 seconds.

Step-by-step explanation:

Use the  motion under gravity  equation s = ut + 0.5gt^2 where u = initial velocity, s = the height, t = time and g = acceleration due to gravity ( 32 fts-2).

So 120 = 95t + 0.5(-32)t^2

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

Solving this using the quadratic formula:

t = [-95 +/- √(95^2 - 4 * -16*-120)] / -32

t = 1.823, 4.1148.

The first value refers to the upward journey and the second to the downward journey.