Yea!
Vf = Vi + at
0 = 20 + -9.8t
t = 2.04s
y2 = Vit + 1/2at^2
y2 = 20(2.04) + 1/2(-9.8)(2.04)^2
y2 = 20.4m
y1(20) + y2(20.4) = yfinal (40.4m)
and the third question is time it took to get there?
Which the initial 2 plus the additional 2.04 = 4.04sec which is on the answer selection...
I'm sorry but terribly confused.
What is time though because to use "final velocity = initial velocity + acceleration * time" to find initial velocity you need time.
Because I tried to find time but it came out weird like -1.03 when I used
Vf = Vi +at
0 = 20 + 9.8t
Sreeja Mobin:
Thanks but how did you get 40m/s.
And for the second equation are all the selected answers incorrect because if that's true I really didn't stand a chance lol.
Okay I got 20 meters from H1 using y = 1/2 (Vf + Vi)t
y = 1/2 (20m/s + 0) 2 = 20m/s
But how would you suggest in getting the second Height.
Would this be right: Vi = 20m/s, Vf = 0, a = 9.8, y=?.
First need time before I could use y = Vit + 1/2 at^2
Vf = Vi +at
0 = 20 + 9.8t
This is where...
Ok I've been trying to do a problem and I can't seem to get it.
A rocket is launched from the origin with an acceleration of 20m/s^2 at an angle of 30.0 degrees above the horizontal. The launch acceleration lasts for 2.00 seconds at which time the fuel is exhausted. The rocket then falls with...