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Dynamics - projectiles.

  1. Jan 16, 2005 #1
    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. jcsd
  3. Jan 16, 2005 #2

    HallsofIvy

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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
     
  4. Jan 16, 2005 #3

    Andrew Mason

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    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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