Question

A ball is thrown into the air with an initial upward velocity of 60ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+60t+6

a. After how many seconds will the ball hit the ground?
b.What will the height be at t=3 seconds
Also, please show work, thanks!

Answer 1

Part a.

You need to solve the quadratic function h = -16t^2 + 60 t + 6 = 0

-16t^2 + 60 t + 6 = 0

Divide by 2, just to make the numbers easier

-8t^2 + 30t + 3 = 0

Use the quadratic formula to find the roots

t=[ -b ± √(b^2-4ac) ] /2a

t=[-30 ± √(30^2-4(-8)(3) ]/2(-8)

t=[-30 ± √(900 + 96) ]/(-16)

t=[-30 ± √(996) ]/(-16)

t=[-30 ± 31.56) ]/(-16)

t=[-61.56) ]/(-16) = 3.85 s

Part b)

h = -16t^2 + 60 t + 6
h (3) = -16(3)^2 + 60 (3) + 6 = -144 + 180 + 6 = 42 ft