If a ball is thrown into the air with an initial velocity of 48 ft/s, then its height (in feet) after t seconds is given by y=48t-16t^2 . Find the velocity when t=2.
2006-09-24 16:41:03 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Take the first derivative of the equation with respect to t.
That gives you dy/dt=48-32t.
Plug in t=2, you get 48-32x2 which equals -16ft/sec.
2006-09-24 16:54:18 · answer #1 · answered by richard Alvarado 4 · 1⤊ 0⤋
The velocity is simply the derivative of the position. Thus v=y'=48-32t, which at t=2 is -16 ft/s.
By the way, I really love it when people posting a question in the mathematics section label their questions as MATH!!!! I mean, what else would it be, philosophy?
2006-09-24 23:46:12 · answer #2 · answered by Pascal 7 · 1⤊ 0⤋
Well, the velocity is just 48 - 32t and at t = 2, it would be -16 ft/s or 16 ft/s downward toward the earth.
2006-09-24 23:51:25 · answer #3 · answered by spongeworthy_us 6 · 1⤊ 0⤋
v = 48 - 32t
then sub t = 2
2006-09-24 23:46:18 · answer #4 · answered by Brian 3 · 0⤊ 0⤋
v = 48 + 32t
t = 2
v(2)=48+64
2006-09-24 23:47:04 · answer #5 · answered by locuaz 7 · 0⤊ 1⤋
ur answer is 32
2006-09-25 00:56:25 · answer #6 · answered by Loser McLoser Face lol 1 · 0⤊ 0⤋