Question

A ball is thrown from an initial height of 4 feet with an initial upward velocity of 40 feet per second. The ball's height h (in feet) after t seconds is given by the following. h=4+40t-16t2 Find all values of for which the ball's height is 26 feet.

Answer 1

Explanation:

To find the times that the height is 26 feet, we set the position equation equal to 26 and solve for t:

`26=-16t^2+40t+4` and

`0=-16t^2+40t-22` and factor that however you are factoring in class to solve a problem like this. When you do that you get

t = .86 seconds and t = 1.68 seconds. That means that .86 seconds after the ball is thrown into the air, it reaches a height of 26 feet; it goes up to its max height and then gravity takes over and pulls it back down. When this happens, it will pass 26 feet again on its way back down. This second time is after 1.68 seconds.