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A particle is projected from a point P (2, 0, 0)m with a velocity 10 m/s making an angle 45º with the horizontal. The line PQ which makes an angle of 37º with positive x-axis lies in the plane of the projectile, xy plane is vertical. The coordinates of the point where the particle will strike the line PQ is: (Take g = 10 m/s2)

  1. (4.5, 15/8, 0)
  2. (8,6,0)
  3. (10,8,0)
  4. (6,10,0)

Rithika , 7 Years ago
Grade 12
anser 1 Answers
Mahesh Koilada
Let the projectile hits line PQ at R(X+2,Y) …........ {assuming that PQ passes through (2,0,0)}
 
Note: here X is measured from (2,0,0)
 
Write kinematic equation for projectile motion in x direction, as below
 
X=V*cosA*t....{X=distance travelled by the projectile in x direction, V=projectile velocity,A=initial angle of projectile, t=time taken to reach point R} …............. (1)
 
Note: g acts only in y direction, x-component of velocity doesn’t change.
 
Y=V*sinA*t + ½*(-g)*t^2 …............... {-ve g is because, Y is +ve in upward direction whereas g acts downwards} .. (2)
 
Slope of the line PQ = tan B = Y/X … (3)
 
  1. Put A=45 deg; devide Y / X and equate it to tan37; we get value of time ‘t’;
  2. Now, substitute the value of ‘t’ in eqn (1) to get X;
  3. Similarly substitute the value of ‘t’ in eqn (2) to get Y;
  4. Finally do not forget to add 2 to X;
  5. You should get X+2 = 4.46 and Y = 1.857
  6. Option A is closely matching;
All the best.
Last Activity: 7 Years ago
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