How Do You Calculate Time of Flight and Range for a Rocket Launched at an Angle?

[HueHue]

Homework Statement

A rocket is launched at an angle of 36.87º (sin=0.6 cos=0.8) with an acceleration of 30m/s² for 20s (fuel runs out). Find a) total time of flight b) horizontal range.

Homework Equations

The Attempt at a Solution

for the first 20 seconds:
v= 30 * 20 = 600m/s d=30 * 20² * 0.5 = 6000m
so vertical distance is= 6000 * 0.6 = 3600m
and horizontal distance is= 6000 * 0.8 = 4800m

fuel runs out
vertical speed is v= 600 * 0.6 = 360m/s
0 = 360 - 10*t .: t=36s (time to reach max height)
max height is= 3600 + 360*36 - 10*36²*0.5 = 10,080m

after reaching max height
10080 = 10 * t² * 0.5 .: t=45s (time to land)
t=36+45=81s (amount of time gravity is doing it's job to make it fall)
horizontal speed is v= 600*0.8 = 480m/s
horizontal distance = 480 * 81 = 38880m

total horizontal distance = 4800 + 38880 = 43680m
total time of flight = 20 + 81 = 101sWell, it's wrong. The answer should be 41,4km and 125s.
Help me please, thanks in advance.

 

[Bystander]

Check the two approaches taken in this thread, https://www.physicsforums.com/threa...th-a-rocket-disagreement.801395/#post-5030959 , and I'll see if I can't get some more help --- looks like it's become a popular problem statement.

 

[RUber]

125 seconds seems to match with 36.87 degrees off of vertical rather than horizontal.
Try the calculations using 0 degrees as vertical.

 

[NascentOxygen]

for the first 20 seconds:
v= 30 * 20 = 600m/s d=30 * 20² * 0.5 = 6000m
so vertical distance is= 6000 * 0.6 = 3600m
and horizontal distance is= 6000 * 0.8 = 4800m

This is saying that its path for the first 20 secs will be a straight line??

 

[RUber]

Since acceleration is given and not thrust, it sounds like this is a straight line simplified problem.

 

[HueHue]

Thanks everyone, I figured out what's wrong. I swaped the 0.6's with the 0.8's
Now I'm getting the right results!
Thanks again, see ya

 

[mfb]

The problem statement is a bit ambiguous in terms of gravity for the first 20 seconds, but as you get the correct result you chose the intended interpretation apparently.

This is saying that its path for the first 20 secs will be a straight line??

Every rocket accelerating from rest will move in a straight line if the direction and magnitude of thrust does not change - constant acceleration from rest gives a straight line.

 

[NascentOxygen]

Every rocket accelerating from rest will move in a straight line if the direction and magnitude of thrust does not change

Indeed, and for fireworks rockets that direction is fixed as "straight out the back" and they invariably follow an arc across the sky. Practically every rocket we observe launched at a low angle does not follow a straight line.

The question could be improved, perhaps by changing to 'a guided missile', so the reader is made aware it is steerable.

 

[mfb]

Practically every rocket we observe launched at a low angle does not follow a straight line.

Usually because thrust or the orientation of the rocket change (e.g. firework rockets have a short acceleration phase and then fly under the influence of gravity), but here the direction and magnitude is fixed, so the rocket follows a straight line until it runs out of fuel.