Lets look at the velocity at 230m
Vf^2=Vi^2+2a*230
solve for Vf (at 230)
vavg= (Vf+51.6)/2
time to go first 230m= 230/vavg
Now time to top...
Vf(top)=Vf(at 230)-gt solve for t
0=Vf(230)-9.81t
so now add this time to time to go first 230. That is the time to get to the max height.
Now , figure the time to fall from the top height. What is the max height?
initial KE+ workby rocket=finaPE
1/2 m 51.6^2+Force*230=mg H
where force= m*a solve for H.
time to fall from H: H=1/2 g t^2 solve for t.
A model rocket is launched straight upward
with an initial speed of 51.6 m/s. It acceler-
ates with a constant upward acceleration of 1.96 m/s2 until its engines stop at an altitude of 230 m. The acceleration of gravity is 9.81m/s^2.
When does the rocket reach maximumheight?
Answer in units of s
How long is the rocket in the air?
Answer in units of s
2 answers
thanks