Respuesta :
v = u + at
= 100 + 30*3 = 190 m/s
lets call the maximum altitude of the rocket H and say that the height reached while the engines are running is h. Also we'll call the angle of the launch a.
we can get an expression for h as follows.
h = vsin(a)t + (1/2)at^2
h = 100*sin(53)*3 + 0.5*30*9
h = 374.6 m
and we can write an expression for the time it takes the rocket to reach maximum height after the engines have failed.
t = [0 - vsin(a)]/-g
= vsin(a)/g
then
H-h = vsin(a)t - 1/2*gt^2
if we sub in the expression for t and rearrange we get
H-h = [(vsin(a)^2]/2g
now add h
H = [(vsin(a)^2]/2g + h
H = 1549 m
b)
to obtain the total time of flight, first work out the time it takes for the rocket to fall from the top of its trajectory.
t = (v-u)/g
where v is now the final vertical velocity and u is the initial vertical velocity.
to get v
v^2 = u^2 + 2gH
so t = [Sqrt(2gH)]/g
now the total time is the falling time plus the time the engines are running, plus the time we calculated previously.
t = [Sqrt(2gH)]/g + 3 + vsin(a)/g
t = 36.72 s
c)
calculating the range is simple since the horizontal velocity stays the same for most of the flight.
We just need to work out the bit of distance traveled while the rocket accelerates for the first 3 seconds of flight.
r = ut + 1/2at^2
r = 100*3+0.5*30*9
r = 435 m
now we take the remaining flight time and put it in the following formula to work out the rest of the distance.
R-r = vcos(a)t
= 190*cos(53)*33.72
= 3856 m
add this to the previous distance we calculated to obtain the range
R = 3856 + 435
R = 4291 m
Hopefully that's right, I may have made an error in there so make sure you check through these calculations yourself to confirm. Hope this helps. :)
= 100 + 30*3 = 190 m/s
lets call the maximum altitude of the rocket H and say that the height reached while the engines are running is h. Also we'll call the angle of the launch a.
we can get an expression for h as follows.
h = vsin(a)t + (1/2)at^2
h = 100*sin(53)*3 + 0.5*30*9
h = 374.6 m
and we can write an expression for the time it takes the rocket to reach maximum height after the engines have failed.
t = [0 - vsin(a)]/-g
= vsin(a)/g
then
H-h = vsin(a)t - 1/2*gt^2
if we sub in the expression for t and rearrange we get
H-h = [(vsin(a)^2]/2g
now add h
H = [(vsin(a)^2]/2g + h
H = 1549 m
b)
to obtain the total time of flight, first work out the time it takes for the rocket to fall from the top of its trajectory.
t = (v-u)/g
where v is now the final vertical velocity and u is the initial vertical velocity.
to get v
v^2 = u^2 + 2gH
so t = [Sqrt(2gH)]/g
now the total time is the falling time plus the time the engines are running, plus the time we calculated previously.
t = [Sqrt(2gH)]/g + 3 + vsin(a)/g
t = 36.72 s
c)
calculating the range is simple since the horizontal velocity stays the same for most of the flight.
We just need to work out the bit of distance traveled while the rocket accelerates for the first 3 seconds of flight.
r = ut + 1/2at^2
r = 100*3+0.5*30*9
r = 435 m
now we take the remaining flight time and put it in the following formula to work out the rest of the distance.
R-r = vcos(a)t
= 190*cos(53)*33.72
= 3856 m
add this to the previous distance we calculated to obtain the range
R = 3856 + 435
R = 4291 m
Hopefully that's right, I may have made an error in there so make sure you check through these calculations yourself to confirm. Hope this helps. :)
