How Do Projectile Dynamics Influence Collision Points?

AI Thread Summary
Projectile dynamics involve the interaction between two particles, A and B, where A is projected upward and B slides down a plane. The calculations for collision points require solving equations for their respective motions, which depend on the angle of projection and initial velocity. The discussion highlights the necessity for both particles to have compatible horizontal velocities for a collision to occur, suggesting a potential contradiction in the problem's setup. Additionally, a separate scenario involves determining the angle needed for a ball to clear a net, requiring similar calculations for trajectory. Overall, the complexities of projectile motion and collision dynamics are emphasized, with solutions hinging on precise mathematical relationships.
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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