SOLUTION: a man jumps off a cliff into water, given the function h(t) = -16t^2+16t+480 where t = time in seconds and h = height in feet At what height did the man jump? What was the the high

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: a man jumps off a cliff into water, given the function h(t) = -16t^2+16t+480 where t = time in seconds and h = height in feet At what height did the man jump? What was the the high      Log On


   

Question 389336: a man jumps off a cliff into water, given the function h(t) = -16t^2+16t+480 where t = time in seconds and h = height in feet At what height did the man jump? What was the the highest point he reached? and how many secs did he take to hit the water?
Found 2 solutions by richard1234, robertb:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The man jumped at 480 feet (t=0, also if you know elementary physics you know that with constant acceleration, the position X is given as %281%2F2%29at%5E2+%2B+%28v_0%29t+%2B+x_0 where x_0 is the original position)

The highest point he reached occurred at the vertex, which is at t = -b/2a = 1/2 second. Plugging in t = 1/2,

h(1/2) = -16(1/4) + 16(1/2) + 480 = 484 feet

To find the amount of time it took to hit the ground, we plug h(t) = 0 and solve:

0+=+-16t%5E2+%2B+16t+%2B+480

I'll divide both sides by -16

0+=+t%5E2+-+t+-+30

0+=+%28t+-+6%29%28t+%2B+5%29

t+=+6 (seconds)


Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The man jumped off the cliff at 480 feet (The height of the cliff, when t = 0.)
The highest point reached is %284ac+-+b%5E2%29%2F%284a%29+=+%284%2A-16%2A480+-+256%29%2F%28-64%29+=+484 feet. (Or 4 feet above cliff level.)
To find the length of time it took him to hit the water, solve for t in
0+=+-16t%5E2+%2B+16t+%2B+480;
0+=+t%5E2+-+t+-+30;
(t-6)(t+5) = 0
The man will hit the water after 6 seconds. (Eliminate the negative value of t.)