A railroad gun of mass M = 2.0 kg fires a shell of mass m=1.0 kg at an angle of θ= 40.0 ∘ with respect to the horizontal as measured relative to the gun. After the firing is complete, the final speed of the projectile relative to the gun (muzzle velocity) is v0=110.0 m/s . The gun recoils with speed vr and the instant the projectile leaves the gun, it makes an angle ϕ with respect to the ground.
What is vp, the speed of the projectile with respect to the ground (in m/s)?
vp=
What is ϕ, the angle that the projectile makes with the horizontal with respect to the ground (in degrees)?
ϕ=
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Its not helping, Can you please give out the formulas without the i' and j' etc??
Thanks
Thanks
Vp = (M*V0*cos(theta))/(M + m) + V0*sin(theta)
Phi = tan_inverse((M+m)*tan(theta)/M)
Phi = tan_inverse((M+m)*tan(theta)/M)
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