Projectile motion problem where the projectile is a rocket

AI Thread Summary
A rocket is launched at a 53° angle with an initial speed of 75 m/s and an acceleration of 25 m/s² for 30 seconds before its engine fails. To find the maximum altitude, total time of flight, and horizontal range, the relevant equations involve resolving the velocity into x and y components and applying kinematic formulas. The maximum altitude can be calculated using the equation y = vsinθt + 0.5at², while the total time of flight and horizontal range require further calculations after the engine failure. The discussion highlights the importance of correctly applying the equations and understanding the transition from powered flight to projectile motion. The problem-solving process emphasizes the need for clarity in defining variables and understanding the physics involved.
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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isTheReau said:

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.
 
I found y using the second equation and got 13,046 m... Would that be ymax? I wasn't sure.
 
isTheReau said:
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
 
okay... and when I solve that, would that be the ymax?
and why are you using -9.81 for the acceleration?
 
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.
 
what is d sub dia?
like... diagonal distance?
 
Okay i think I get it now... thank you!
 
isTheReau said:
what is d sub dia?
like... diagonal distance?

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