a ball is thrown vertically upward w/ an initial velocity of 80 ft/sec. the distance "s" (in ft) of the ball from the ground after "t" sec is s=80t-16t^2.
how do i figure out the time interval when the ball is more than 96 ft above the ground?
and how many sec it takes to reach the max height of the ball?
an explanation would be helpful and thanks for your help.
2006-09-20 10:47:46 · 2 answers · asked by shih rips 6 in Science & Mathematics ➔ Physics
First, find out when the ball is exactly 96 feet above the ground. To get this, plug in 96 feet for "s" and solve:
96 = 80t - 16t^2
Divide through by 16 and rearrange to get:
t^2 -5t + 6 = 0
This factors nicely into (t-3)*(t-2) = 0
Hence, the equation is true when t=2 or t=3.
The ball is 96 feet above the ground between t=2 and t=3 seconds.
Since parabolas are symmetrical, the maximum height would be at t=2.5 seconds.
2006-09-20 10:55:08 · answer #1 · answered by Bramblyspam 7 · 0⤊ 0⤋
1) solve the eq 96 = 80t -16t^2 for t. Hopefully you will get 2 answers, which will be the 2 times when the ball is 96 ft high.
2) When the ball is highest, it's velocity (ds/dt) is zero; solve
ds/dt = 0 = 80 - 32t for t and the answer is yours.....
2006-09-20 17:55:26 · answer #2 · answered by Steve 7 · 0⤊ 0⤋