A ball is thrown vertically upward with an initially velocity of 48 feet per second. the distance s(in feet) of the ball from the ground after two seconds is s=48t-16t^2. For what time t is the ball more than 20 feet above the ground?

Question

cathyangs · Answer

for .5 seconds to 2.5 seconds.

anonymous · Answer

48t - 16t^2 > 20 12t - 4t^2 > 5 4t^2 - 12t + 5 > 0 4t^2 - 10t - 2t + 5 > 0 2t(2t - 5) - (2t - 5) > 0 (2t - 5)(2t - 1) > 0 hence 0 <= t < 0.5 or t > 2.5

anonymous · Answer

haha, oops, i flipped the equality sign,

so yes it should be 0.5 <= t <= 2.5

anonymous · Answer

errr, 0.5 < t < 2.5