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# A shot is fired at a distance of 39.2m from the foot of a pole 19 m high so that is just passes over it. Find the magnitude and direction of the velocity of the shot.

Abhishek Pachauri
35 Points
6 years ago
the magnitude of velocity is 28m/s and direction is 45 degree from horizontal.  THANK U
Sdk
11 Points
4 years ago
Let D be the point of projection and AC be the pole and B be the point where it reaches the ground.Then,CD=39.2BD=2*39.2=78.4Given that the body passes over the vertical pole 19.6mThenTaking downward motion from A to Bt=root of (2h/g)=root of (2*19.6/9.8)=2sTaking motion from D to a to BTime of flight=(2usinN) /g N (angle of projection)time of ascent=time of descent=usinN/g=usinN=19.6Taking horizontal motion from D to BucosN*Time of flight=78.4ucosN=78.4/2*2=19.6Then usinN/ucosN=19.6/19.6=1implies tanN=1then N=45We know range is maximum when N =45Then u^2sin2N/g=78.4Solving we get u= 27.7m/s