An object is thrown upward from a height of 24ft. Height is h of the object t sec after the object is released given by h=-16t^2+24t +24

Question

How long does it take the object to reach a height of 8 feet?

How long does it take the object to hit the ground?

Answer 1

When t=0, h= 24' which is the starting point, 24' above ground.

1. When h=8' above the ground h=8=-16t^2+24t+24. Therefore, dividing by 8 and rearranging we have 2t^2-3t-2=0=(2t+1)(t-2), so t=2 is the only positive root, making it 2 seconds to reach a height of 8' above ground.
2. To fall to the ground h=0=-16t^2+24t+24 to 2t^2-3t-3=0=t^2-3t/2-3/2. Completing the square we have t^2-3t/2+9/4=3/2+9/4 and (t-3/2)=±√(3/2+9/4)=±√15/2. Therefore, time to reach the ground=3/2+√15/2 (the only positive root)=3.44 seconds approximately.