A 250 kg projectile is fired from the ground with a speed of 120m/s at a 60degre

Question

A 250 kg projectile is fired from the ground with a speed of 120m/s at a 60degree angle over the horizontal.At the highest point of its arc it explodes into three pieces A,Band C.Immediately after the explosion pieces A and B are still moving with the same speed as the whole projectile had right before the explosion.However,A moving vertically downward while B is moving horizontally forward ( same direction as the projectile prior to its explosion).find the velocity of C right after the explosion.

Explanation / Answer

you first use the projectile motion equations to find the velocity at the peak x=Vx*t where Vx=120cos60 y=Vy0*t-(1/2)gt^2 V0y the initial y velocity is 120sin60. Vy=120sin60-gt at the point where it splits the y velocity is 0 since it is at the top of the arc, and the x velocity is constant 120cos60, so this is the overall velocity. since A and B are the same speed as the projectile the velocities are this same magnitude in the specified directions. then you use conservation of momentum in the x and y direction. mass times velocity for each piece should be the same overall in both x and y. piece A is going straight down, so it has momentum of 1/3 mass times velocity in the negative y direction. piece b is moving horizontally so it has no y momentum. since the y momentum before the split was 0 the y momentum in c has to cancel the y momentum in a, so the same m*v as a but in the opposite direction, positive y. then do the same thing for the x direction to get the x velocity of c. then add vectors to get the final velocity of c. i am sure this will help you finsih the problem please let me know if you need any further clarifications and please dont forget to rate my answer :)

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