Question

HELPPPPP!! A juggler tosses a ball into the air. The ball leaves the juggler's hand 5 feet above the ground and has an initial velocity of 30 feet per second. The juggler catches the ball when it falls back to a height of 2 feet. How long is the ball in the air? Use the model h(t)=−16t2 +v0t +h0, where v0 is the initial velocity (in feet per second), h is the height (in feet), t is the time in motion (in seconds) and h0 is the initial height (in feet).
Remember to round your answer to the nearest hundredth.

Answer 1

Answer:

t = 2.525 seconds

Step-by-step explanation:

h(t) = -16t^2 + v*t + h0  h(t) = the height above the ground after t seconds v = 40 ft/s initial velocity  h0 = 4 ft the initial height of the object (the ball)  t = the time of the motion  h(t) = -16t^2 + 40*t + 4  we need to find t such that h(t) = 3 ft  3 = -16t^2 + 40*t + 4  -16t^2 + 40*t + 4 = 3  by solving we find

t = 2.525 seconds

You can round on your own.