Finding Maximum Altitude of a Rocket

[gneill]

Have you studied projectile motion, where the horizontal and vertical components of the motion are treated independently?

The velocity vector for a projectile, at any given instant, is made up of two components: a vertical component and a horizontal component. Only the vertical component of the motion is important for calculating the height of the projectile at any time.

In calculating the height of the rocket projectile, you want to concern yourself only with the vertical component of the velocity. That is the 1380*sin(70°) value.

 

[stupif]

ya...i have studied.
what i know in projectile motion is vetical velocity in changing because of the gravity,use this formula, s=ut +1/2 at^2,
the horizontal velocity is constant because of the inertia , so use s=vt

so, 1380sin 70 = 1296.78
after that, what should i do?

 

[gneill]

ya...i have studied.
what i know in projectile motion is vetical velocity in changing because of the gravity,use this formula, s=ut +1/2 at^2,
the horizontal velocity is constant because of the inertia , so use s=vt

so, 1380sin 70 = 1296.78
after that, what should i do?

1296.78 m/s is the "u" in your u*t + 1/2 at² formula. It's the initial vertical velocity at the beginning of the free-fall portion of the rocket's trajectory.

But you want to find the maximum height. So you might want to choose another formula to use with this velocity. Previously you wrote the equation, v_f² = v_i² + 2*a*d. Here the initial velocity, v_i, would be your u.

 

[stupif]

so is 19451.64m + 85797.88= 105249.52m
but the answer is 105.6km. a slightly differences. is it correct??

 

[gneill]

so is 19451.64m + 85797.88= 105249.52m
but the answer is 105.6km. a slightly differences. is it correct??

The difference is probably due to the use of slightly difference value for g, the acceleration due to gravity. But yes, the value looks okay.

 

[stupif]

okay~thank you very much. now the next question is the time of flight from launching to impact...answer is 308.65s
i don't know what the question talking about?
what is the impact it means?

 

[gneill]

okay~thank you very much. now the next question is the time of flight from launching to impact...answer is 308.65s
i don't know what the question talking about?
what is the impact it means?

Presumably that would be when the rocket crash-lands on the ground. After the fuel was used up the rocket became a free-falling projectile. It will rise to its maximum height and then fall back to the ground.

 

[stupif]

thanks, i got it, now the next question is the distance from launched pad to impact point.
the answer is 139km

1st i find the distance at s1, 20700cos 70 = 7029.82m
2nd, i find the distance between s1 and s2 , s=vt, s= 1380cos70 X 28.19s = 13305m
3rd, i find the distance between s2 to the ground, s=vt , s= 1380cos70 X 146.50s =69146m

total up is 89480.82m

i think something wrong in step 2 and 3.

 

[gneill]

It's difficult to comment without knowing where the numbers and constants are coming from. What are s1 and s2? Where do the time intervals 28.19s and 146.50s come from?

 

[stupif]

s1 is the distance from rest to h1
s2 is the distance from h1 to h2 when the engines shut off

mistake for 28.19. it should 30s
146.50 is come from this equation, s = ut +1/2at^2 , 105.16km= 0t+ 1/2(9.8)t^2
t= 146. 50s

 

[gneill]

s1 is the distance from rest to h1
s2 is the distance from h1 to h2 when the engines shut off

mistake for 28.19. it should 30s
146.50 is come from this equation, s = ut +1/2at^2 , 105.16km= 0t+ 1/2(9.8)t^2
t= 146. 50s

It looks as though your s1 and s2 values are horizontal distances, with s1 being the the horizontal distance from launch to where the fuel runs out, and s2 the horizontal distance from there to the maximum height.

Why do you take the time between s1 and s2 to be 30 seconds?

 

[stupif]

sorry...mistake again...i know how to do already, just now made some mistakes in calculation.
thank you very much...