A rock is thrown in the air at time t=0. It's height h(t) in feet above the ground at time t in seconds is given by h(t)=-4t^2+24t+5. At what time (s) is the height of the rock 5 feet above the ground?

Question

saifoo.khan · Answer

h(t)=-4t^2+24t+5

Now since the height is 5,

5 = -4t^2+24t+5

Solve for "t".

saifoo.khan · Answer

You''be left over with:

-4t^2+24t = 0

Now solve for t.

anonymous · Answer

it means tht......
5=-4t^2+24t+5
Subtrating by 5 on both sides,
we get
-4t^2+24t=0
4t is common here so by factorizing
we get
-4t(t-6)=0
therefore -4t=0 or t-6=0
                t=0(rejected) or t=6
the time taken =6

anonymous · Answer

Thank you, I thought I was going crazy with this problem. It's great to know that I was not too far off from the right path. Thank you, you all are life savers!!!!!!!

anonymous · Answer

Its ok we r here to help you!!!!!