Attempted Part I & II Problems

  • Thread starter Thread starter theredbarronmvr
  • Start date Start date
AI Thread Summary
The discussion centers on solving parts i and ii of a physics problem related to projectile motion. The user is struggling with part ii, specifically how to reach the x value found in part i without an air resistance term. It is clarified that without air resistance, the problem simplifies to basic projectile motion. Additionally, a key point is made regarding the evaluation of an integral, emphasizing that the constant c2 should be set to zero before integration to avoid invalid assumptions. The conversation highlights the importance of correctly applying mathematical principles in physics problems.
theredbarronmvr
Messages
2
Reaction score
0
Homework Statement
A bullet of mass m is shot vertically up with initial speed v0. It is subject to air resistance that is quadratic in speed (constant of proportionality c2).

i. What is the maximum height reached?

ii. What initial speed u0, would be necessary to reach the same maximum height in the absence of air resistance?

Express your answers in terms of c2, g, m, and v0.
Relevant Equations
For part i: m(dv/dt) = -c2(v^2) - mg
I have an attempted solution to part i here. I'm primarily having trouble with part ii since I don't see how the x found in part i can be attained if there's no air resistance term.
 

Attachments

  • pf2.jpg
    pf2.jpg
    29.4 KB · Views: 119
Physics news on Phys.org
theredbarronmvr said:
I'm primarily having trouble with part ii since I don't see how the x found in part i can be attained if there's no air resistance term.
If there is no air resistance, then it's a simple projectile motion problem.
 
  • Like
Likes theredbarronmvr
PeroK said:
If there is no air resistance, then it's a simple projectile motion problem.
Ahh of course, thank you! I was definitely overthinking it.
 
theredbarronmvr said:
I have an attempted solution to part i here. I'm primarily having trouble with part ii since I don't see how the x found in part i can be attained if there's no air resistance term.
In case you're wondering, when you evaluated the integral, you assumed ##c_2 \ne 0##, so it isn't valid to set ##c_2=0## in the expression you derived. (I assume that's the difficulty you were running into.) You have to set ##c_2## to 0 before you integrate.
 
  • Like
Likes theredbarronmvr
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top