A ball is thrown into the air with an initial upward velocity of 60 ft/s. Use the equation h=-16t^2+60t+6 to answer.

Question

How long will it take for the ball to hit the ground (when h=0)?

When will the ball reach its maximum height?

What is the maximum height of the ball?

What is the height of the ball after 3 seconds?

Answer 1

h(t)=0 = -16t^2 +60t + 6

-4t^2 +15t +1.5 =0

t = 15/8 + (1/8)sqr(225+24) = 1.875 +(sqr249)/8 = about 3.85 seconds. It hits the ground a little over twice the time to reach the maximum height. Had it started at ground level, it would have been exactly twice the time for maximum height.

maximum height is when t= 1.875 seconds

h'(t) = -32t + 60=0, t = 60/32 = 15/8 = 1.875 seconds

maximum height = -16(15/8)^2 + 60(15/8) + 6 = (1/8)(-450+900) + 6 = 450/8 + 6 = 56.25 + 6 = 62 1/4 feet

height after 3 seconds is

h(3) =-16(3)^2 + 60(3) + 6 = 42 feet

this parabola equation reflect an initial height of 6 feet, an initial velocity of 60 feet per second and gravity of -32 feet per second per second = deceleration.