Projectile motion problem where the projectile is a rocket

  • Thread starter isTheReau
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  • #1
isTheReau
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Homework Statement


A rocket is launched at an angle of 53° above the horizontal with an initial speed of 75 m/s. It moves in powered flight along its initial line of motion with an acceleration of 25 m/s^2. At 30 s(seconds) from launch, its engine fails and the rocket continues to move as a projectile.
(a) What is the rocket's max altitude? (ymax)
(b) What is the rocket's total time of flight? (tτ)
(c) What is the rocket's horizontal range? (R/x)

My given information looks like this: θ= 53°, v1= 75 m/s, a=25 m/s, t= 30s, Δy=0


Homework Equations


x=vcosθt
y=vsinθt+.5at^2
ymax= -(vsinθ)^2/2a

The Attempt at a Solution


I tried resolving the velocity into x and y components but I'm not even sure if that's useful information. I also attempted to use the above equations but to no avail... I just need to get this started so that I can finish it. The problem is I'm not sure where to start.
 
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Answers and Replies

  • #2
NasuSama
326
3

Homework Statement


A rocket is launched at an angle of 53° above the horizontal with an initial speed of 75 m/s. It moves in powered flight along its initial line of motion with an acceleration of 25 m/s^2. At 30 s(seconds) from launch, its engine fails and the rocket continues to move as a projectile.
(a) What is the rocket's max altitude? (ymax)
(b) What is the rocket's total time of flight? (tτ)
(c) What is the rocket's horizontal range? (R/x)

My given information looks like this: θ= 53°, v1= 75 m/s, a=25 m/s, t= 30s, Δy=0


Homework Equations


x=vcosθt
y=vsinθt+.5at^2
ymax= -(vsinθ)^2/2a

The Attempt at a Solution


I tried resolving the velocity into x and y components but I'm not even sure if that's useful information. I also attempted to use the above equations but to no avail... I just need to get this started so that I can finish it. The problem is I'm not sure where to start.

Yes, these formulas are useful to determine the certain components. To find the altitude, you need to use the second formula: y=vsinθt+.5at^2

Substitute the given values and try to find the max. You can perform the differentiation if you know calculus.

dy/dt = vsinθ + at

Then, let dy/dt = 0 and solve for t. This gives you the extrema if a = g = -9.81.
 
  • #3
isTheReau
5
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I found y using the second equation and got 13,046 m... Would that be ymax? I wasn't sure.
 
  • #4
NasuSama
326
3
I found y using the second equation and got 13,046 m... Would that be ymax? I wasn't sure.

Actually not. You actually solve for t for the dy/dt equation where dy/dt = 0, so we have...

0 = vsinθ + at
t = -vsinθ/a

Now, find the value of t via substitutions. Know that:

θ = 53°
v = 75
a = g = -9.81

Finally, substitute the t value for y = vsinθt + ½at²
 
  • #5
isTheReau
5
0
okay... and when I solve that, would that be the ymax?
and why are you using -9.81 for the acceleration?
 
  • #6
azizlwl
1,065
10
A rocket is launched at an angle of 53° above the horizontal with an initial speed of 75 m/s. It moves in powered flight along its initial line of motion with an acceleration of 25 m/s^2. At 30 s(seconds) from launch, its engine fails and the rocket continues to move as a projectile.
(a) What is the rocket's max altitude? (ymax)
(b) What is the rocket's total time of flight? (tτ)
(c) What is the rocket's horizontal range? (R/x)

My given information looks like this: θ= 53°, v1= 75 m/s, a=25 m/s, t= 30s, Δy=0

------------------------------------
Given θ= 53°, v0= 75 m/s, a=25 m/s2, t= 30s

First part of flight
v1-v0=at
v1=825m/s
ddia=0.5x825x30=12375m

2nd part of flight.
Given θ= 53°, v0= 825 m/s, a=-9.8 m/s2, v1=0

Maximum height is maximum height of 2nd. part plus y component of ddia.
 
  • #7
isTheReau
5
0
what is d sub dia?
like... diagonal distance?
 
  • #8
isTheReau
5
0
Okay i think I get it now... thank you!
 
  • #9
azizlwl
1,065
10
what is d sub dia?
like... diagonal distance?

Yes
Sorry, i should have defined it.
Anyway you got it.
 

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