A ball is thrown from an initial height of 5 feet with an initial upward velocity of 36 ft/s

Question

A ball is thrown from an initial height of 5 feet with an initial upward velocity of 36 ft/s. The ball's height h (in feet) after t seconds is given by the following. h= 5+36t-16^2.

Find all values of t for which the balls height is 17 feet.

Round your answer to the nearest Hundredth.

Answer 1

h(t) = -16t^2 +36t +5 = 17

-16t^2 +36t -12 = 0

16t^2 -36t -+12 = 0

4t^2 -9t +3 = 0

t = 9/8 + or - (1/8)sqr(81 -48) = about 1.84 and 0,41 seconds when it reaches 17 feet.

0.41 seconds on the way up is the first time it reaches 17 feet. Then it . Hits a maximum height in 1.125 seconds, reaching about 21 feet, then at 1.8 seconds it's at 17 feet again as it falls back down

graph it, it's a downward opening parabola with vertex (1.125, 20.75) = maximum point