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 v
elocity 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.
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