Question

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 1

Answer:

Mag of velocity = 8.85 m/s

Direction = 26.56° from horizontal

Explanation:

Find the angle by making right angled triangle and using tan

then use the max height formula to find the velocity

Answer 2

Velocity of the Shot is 28 m/s

Explanation:

• Let D be the point of projection and AC be the pole and B be the point where it reaches the ground
• Then,CD=39.2, BD=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=us in N/g=usinN=19.6
• Taking horizontal motion from D to Bucos N*Time of flight=
• 78.4 ucosN=78.4/2*2=19.6
• Then  
• usinN/ucosN=19.6/19.6=1 implies tan N=1 then N=45
• We know range is maximum when N =45  Then,  
• $u^2sin2N$/g=78.4
• Solving we get  u= 27.7m/s , (means Answer is 28 m/s)