1.

a shot is fired at a distance of 39.2m from the foot of a pole 19.6m high so that it just passes over it. Find the magnitude and direction of the velocity of the shot

Answer»

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=78.4

Given that the body passes over the vertical pole 19.6m

Then Taking downward motion from A to Bt=ROOT of (2h/g)=root of (2*19.6/9.8)=2s

Taking motion from D to a to B

TIME of flight=(2usinN) /g N (angle of projection)

time of ascent=time of descent=usinN/g=usinN=19.6

Taking horizontal motion from D to BucosN*Time of flight=78.4ucosN=78.4/2*2=19.6

Then usinN/ucosN=19.6/19.6=1implies tanN=1

then N=45We know RANGE is maximum when N =45

Then u^2sin2N/g=78.4

Solving we get u= 27.7m/s


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