Question

A ball is thrown from an initial height of 7 feet with an initial upward velocity of 23 ft/s. The ball's height h (in feet) after 1 seconds is given by the following.
h = 7+23t-16t^2
Find all values of 1 for which the ball's height is 15 feet.

Answer 1

Answer:

Step-by-step explanation:

If we are looking for the time(s) that the ball is at a height of 15, we simply sub in a 15 for the height in the position equation and solve for t:

$15=-16t^2+23t+7$ and

$0=-16t^2+23t-8$

Factor this however you factor a quadratic in class to get

t = .59 seconds and t = .85 seconds.

This means that .59 seconds after the ball was thrown into the air it was 15 feet off the ground. Then the ball reached its max height, gravity took over, and began pulling it back down to earth. The ball passes the height of 15 feet again on its way down after .85 seconds.