Question 314607: Jack throws a ball straight up in the air. The equation, h(t)= -16tsquared+64t+5 gives the height (in feet) of the ball from the ground t seconds after it is thrown. How long does it take for the ball to hit the ground?
I have already figured out the maximum height, I think, as 75 feet. Which takes 5 seconds.
Thanks.
Found 2 solutions by texttutoring, malaydassharma:
Answer by texttutoring(324)   (Show Source):
You can put this solution on YOUR website! I think that the max height is 69 feet, and it occurs at 2 seconds.
Make sure you complete the square:
h(t) = -16t^2+64t+5
= -16(t^2 - 4t) + 5
=-16(t^2 -4t + 4) + 5 + 64
=-16(t-2)^2 + 69
This a parabola with vertex at (2,69).
If you want to find the time that it takes to hit the ground, set h=0 and use the quadratic formula to solve for t.
0=-16t^2+64t+5
The quadratic formula should give you two answers: t=-0.77 and t=4.077
You have to choose the positive answer, as time cannot be negative.
Therefore, the ball hits the ground at 4.077 seconds.
Answer by malaydassharma(59)  (Show Source):
You can put this solution on YOUR website! When ball hits the ground, h(t)=0. Hence,
h(t)=-16t^2+64t+5=0
or 16t^2 -64t -5=0 , a simple quadratic eqn.
Putting the appropriate values in the standard solution of a Quadratic Equation
 , we get ,
t = [ - ( - 64) + - sqrt { ( - 64)^2 - 4*16*( - 5)} ] / (2 * 16)
t = [ 64 + - sqrt { 64 ( 64 + 5} ] / (2 * 16)
t = [ 64 + - 8 * sqrt ( 69) ] / (2 * 16)
t = [ 8 + - sqrt ( 69) ] / 4
As t can not be negative and sqrt(69) > 8 , so the only acceptable solution for t is
t = [ 8 + sqrt ( 69) ] / 4
t=16.31 sec.
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