# Homework Help: Dynamics - projectiles.

1. Jan 16, 2005

### fasterthanjoao

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.

Last edited: Jan 16, 2005
2. Jan 16, 2005

### HallsofIvy

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(&theta;), y(t)= -(g/2)t2-Vsin(&theta;).

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

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

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

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

Solve those equations for t and &theta;

3. Jan 16, 2005

### Andrew Mason

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