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Janell is standing on a set of bleachers and throws a ball into the air at an initial velocity of

35 ft/s. The height of the ball, h, at 1 seconds is modeled by the equation h = -162 +35t + 6



How many seconds will it take the ball to reach the ground

1 Answer

1 vote

Answer:

The time taken to reach ground is 3.45 sec

Explanation:

Given as :

Distance of ball = h = -162 + 35t + 6 at t = 1 sec

The speed of ball thrown = 35 ft per sec

Let the Time taken to reach ground = T sec

So Time =
(Distance)/(Speed)

Or, Time =
(-162+35t+6)/(35)

Now distance h at t= 1 sec ,

h = -162 + 35t +6

or, h = - 162 + 35(1) + 6

So, h = - 121 feet

Or
\begin{vmatrix}h\end{vmatrix} = 121 feet

Or, Time =
( 121)/(35)

Or, Time = 3.45 sec

Hence The time taken to reach ground is 3.45 sec Answer

User Santhosh Gandhe
by
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