How Do Projectile Dynamics Influence Collision Points?

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SUMMARY

The discussion centers on the dynamics of projectile motion involving two particles, A and B. Particle A is projected from a height with an initial velocity V at an angle theta, while particle B slides down a smooth inclined plane at the same angle. The equations of motion for both particles are derived, leading to the conclusion that for a collision to occur, their horizontal velocities must align, which presents a contradiction in the initial conditions. The second scenario involves a ball projected from a height of 2m with a speed of 16m/s, requiring calculations to determine the angle needed to clear a 1m high net located 10m away.

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fasterthanjoao
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1) A particle A, projected upward with velocity V in a direction making an angle theta with the horizontal, is fired from the top of a smooth plane whose angle of inclination to the horizontal is also theta. At the moment of projection a second particle B slides from rest at the point of projection. At the end of its flight projectile A strikes particle B. Calculate angle of projection and time of flight.


-beats me, if you know how to do it a rough outline will be fine.


2) A ball is porjected from a point 2m high with speed 16m/s. what should be the angle to clear a net 1m high, distant 10m from the point of projection?

thanks.
 
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The only way (1) makes sense to me is if particle A is fired "to the left" while the plane is "to the right". Taking its initial positon (the "point of projection") its position at time t is given by x(t)= -Vcos(θ), y(t)= -(g/2)t2-Vsin(θ).

Particle B, sliding along the plane has acceleration -gsin(θ)cos(θ) in the x direction, -gcos2(θ) in the y direction. Its position at time t is given by x(t)= -(g/2) sin(θ)cos(θ)t2, y(t)= (-g/2)cos2(θ)t2.

In order that the two particles collide, we must have
x(t)= -Vcos(θ)= -(g/2) sin(θ)cos(θ)t2 and
y(t)= -(g/2)t2-Vsin(θ)= (-g/2)cos2(θ)t2.

Solve those two equations for t and θ (the answer will depend upon V).

2) The equations of motion of the ball are x(t)= 16cos(θ)t, y(t)= (-g/2)t2+ 16sin(θ)+ 2. In order to clear the net, we must have y at least 1 when x= 10. That is:
10= 16cos(θ)t, 1= (-g/2)t2+ 16sin(θ)+ 2

Solve those equations for t and θ
 
fasterthanjoao said:
1) A particle A, projected upward with velocity V in a direction making an angle theta with the horizontal, is fired from the top of a smooth plane whose angle of inclination to the horizontal is also theta. At the moment of projection a second particle B slides from rest at the point of projection. At the end of its flight projectile A strikes particle B. Calculate angle of projection and time of flight.
The only way that A can hit B is if B and A have horizontal velocity in the same direction. But this contradicts the question, so there is no solution. Perhaps they mean the plane has an angle with the horizontal of pi-theta.

AM
 

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