Projectile motion problem where the projectile is a rocketby isTheReau Tags: help asap, motion, projectile, rocket 

#1
Sep2312, 05:20 PM

P: 5

1. The problem statement, all variables and given/known data
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 2. Relevant equations x=vcosθt y=vsinθt+.5at^2 ymax= (vsinθ)^2/2a 3. 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. 



#2
Sep2312, 05:49 PM

P: 326

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
Sep2312, 06:00 PM

P: 5

I found y using the second equation and got 13,046 m... Would that be ymax? I wasn't sure.




#4
Sep2312, 06:20 PM

P: 326

Projectile motion problem where the projectile is a rocket0 = 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
Sep2312, 08:00 PM

P: 5

okay... and when I solve that, would that be the ymax?
and why are you using 9.81 for the acceleration? 



#6
Sep2312, 08:57 PM

P: 961

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/s^{2}, t= 30s First part of flight v1v0=at v1=825m/s d_{dia}=0.5x825x30=12375m 2nd part of flight. Given θ= 53°, v0= 825 m/s, a=9.8 m/s^{2}, v1=0 Maximum height is maximum height of 2nd. part plus y component of d_{dia}. 



#7
Sep2312, 09:12 PM

P: 5

what is d sub dia?
like... diagonal distance? 



#8
Sep2312, 09:18 PM

P: 5

Okay i think I get it now... thank you!




#9
Sep2312, 09:21 PM

P: 961

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


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