A ball is thrown upward into the air with an initial velocity of 64ft/s. If H gives the height of the ball at time t, then the equation relating h and t is
h(t) = -16t2(the 2 is squared) +64t
Find the maximum height the ball will attain.
2006-07-26 16:10:43 · 5 answers · asked by brittneyafoster 1 in Education & Reference ➔ Higher Education (University +)
Vertex first: x= -b/2a = -64/[2(-16)] = 2
h(2) = -16*2^2 +64*2 = -64+128 = 64 ft
2006-07-26 16:18:29 · answer #1 · answered by jenh42002 7 · 0⤊ 0⤋
This is more of a high school physic question.
Just to give you some tips:
The ball is thrown upward, so the velocity is going to decrease as the ball is going up. Eventually, the ball will reach its maximum height and start falling back down. At the moment when the ball reach its maximum height, the velocity is zero.
Yes, the person below my answer gave you the correct answer or way to solve this problem.
2006-07-26 23:16:12 · answer #2 · answered by ? 5 · 0⤊ 0⤋
Alternatively I like this elegant way...
h(t)= -16t^2+64t
h(t)=-16(t^2-4)
h(t)=-16((t-2)^2-4)
h(t)=-16(t-2)+64
This is a max when -16(t-2)=0 so when t=2 and the height is then
-0+64=64feet
See completing the square has practical uses!
2006-07-27 00:04:34 · answer #3 · answered by dope_move_busta 1 · 0⤊ 0⤋
since this question has been posted in Higher education, im assuming you know calculus.
get the first derivative of the equation and equate it to 0 find the time (t) and plug it back to the original equation
after the first derivative the equation becomes -32t +64
equate to 0 to give you t = 2.
substitute t =2 to the original h(t) equation to get the answer.
2006-07-26 23:16:52 · answer #4 · answered by aking 2 · 0⤊ 0⤋
You need to learn to do your own homework.
2006-07-26 23:15:02 · answer #5 · answered by Cabana C 4 · 0⤊ 0⤋