a rock is thrown vertically from the ground with a velocity of 24 meters per second, and it reaches a height of 2+24t-4.9t^2 after t seconds. How many seconds after the rock is thrown will it reach max. height and what is the max. height of the rock in meters? how many seconds after it is thrown will it hit the ground?
2007-10-28 05:37:21 · 3 answers · asked by ~USA~ 2 in Science & Mathematics ➔ Mathematics
max height occurs when t = -b/2a = -24/ -9.8 = 2.45 seconds
Thus max height = 2+24(2.45) -4.9(2.45)^2=31.39 m.
It hits ground when height = 0
So -4.9t^2 +24t +2 = 0
t = [-24 +/- sqrt(24^2-4(-4.9)(2))]/(-4.9*2)
t = [-24 +/- sqrt(615.2)]/-9.8
t = [-24 +/- 24.8]/-9.8
t = - 48.8/-9.8 = 4.98 seconds
2007-10-28 06:07:45 · answer #1 · answered by ironduke8159 7 · 1⤊ 0⤋
You have h(t)=2+24t-4.9t^2
Then h'(t)=24-9.8t
The max. of h(t) occurs
where h'(t)=0 -->t=24/9.8 sec.
The max height is
h(24/9.8)=.....
It will hit the ground
in 2*24/9.8=24/4.9 sec.
The information for the velocity
is not necessary because it
can be found
v(t)=h'(t)=24-9.8t
The initial velocity is v(0)=24 m/s
2007-10-28 06:08:27 · answer #2 · answered by katsaounisvagelis 5 · 0⤊ 0⤋
I don'tknow dude, I'll smoke another bowl and think about it.
2007-10-28 06:07:35 · answer #3 · answered by fixn2rock 2 · 0⤊ 1⤋