Question 1358: How do I find the height of this quadratic word problem?
If a ball is thrown straight up with an initial velocity of 48 feet per second, its height s after t seconds is given by the equation s=48t-16tsquared. Find the maximum height attained by the ball and the time it takes for the ball to return to earth.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Apply complete square to the right of s(t), we have
Since s(t) = 48 t-16t^2 = -16(t^2 -3t + (3/2)^2) + 16(3/2)^2
= -16(t - 3/2)^2 + 36
We know -16(t - 3/2)^2 <= 0 for all t.
So, s(t) <= 36 and when t = 3/2 ,s(t) attains its maximum value 36.
Then to find how long it takes the ball returning the ground, we
have to solve s(t) = -16 t(t-3) = 0, we get t =3 or 0(inital).
[Or since we know that it takes 3/2 sec for the ball to reach its max height,
then it takes 2*3/2 =3 sec to return to earth.]
Hence, the maximum height of the ball is 36 ft and it takes 3 sec
for the ball to return to earth .
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