SOLUTION: The height of a ball after being dropped from a point 100 ft above the ground is given by: h(t)=-16t^2+100. When will the ball be 60ft above the ground? What is the max. height r

Question 284708: The height of a ball after being dropped from a point 100 ft above the ground is given by: h(t)=-16t^2+100.
When will the ball be 60ft above the ground?
What is the max. height reached?
When will the ball reach the ground?
Answer by toidayma(44) (Show Source): You can put this solution on YOUR website!
h(t) = -16t^2 + 100
First we see that at t0 = 0, the ball is at the height h(t0) = h(0) = 100 (ft)
A> Let t1 (seconds)is the time needed for the ball to fall from the height h(t0) = 100ft to h(t1) = 60ft. Of course t1 >0. We have:
h(t1) = -16t1^2 + 100 = 60 <-> 40 = 16*t1^2 = 40 <-> t1^2 = 2.5 <-> t1 = 1.25(s) (You can get -1.25(s) since t1 > 0.)
Thus, After 1.25 second, the ball will fall from 100ft to 60ft above the ground.
B>What is the max height reached?
This means you have to find the maximum value of h(t)
You can easily see that h(t) = -16t^2 + 100 =< 100 since -16t^2 >= 0.
And "=" occurs when -16 t^2 = 0 <-> t = 0 = t0.
And at t0, the maximum height that the ball reached is h(t0) = h(0) = 100ft.
C> This means you have to solve the equation h(t) = 0
Let t2 (seconds) is the time needed for the ball to fall from the height h(t0) to the height h(t2) = 0(ft) (i.e the ball reaches the ground). Once again, of course t2> 0. You have
h(t2) = 100 -16*t2^2 = 0 <-> 16*t2^2 = 100 <-> t2^2 = 6.25 <-> t2 = 2.5 (seconds)
Thus, After 2.5 seconds, the ball will fall from 100ft down to the ground.
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