Question

A ball is thrown into the air at an initial velocity of 18 meters per second from an initial height of 10 meters. The equation that models the path of the ball is given by h=-4.9t^2+18t+10h How high is the ball at 3 seconds?
Group of answer choices

19.9 meters

108.1 meters

10 meters

The ball is no longer in the air at 3 seconds.

Answer 1

Answer:

Step-by-step explanation:

This is your position equation:

$s(t)=-4.9t^2+18t+10$

There's a whole lot of information in that equation, but what we are concerned about right now is the height of the ball after t = 3 seconds.  If this is the position of the ball at any time t, we will sub in 3 for t to find out where the ball is at 3 seconds.

$s(3)=-4.9(3^2)+18(3)+10$

which simplifies to

s(3) = -44.1 + 54 + 10 which is

s(3) = 19.9 meters

That's how high the ball is in the air at 3 seconds.