Question

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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 45∫ with the horizontal. The plane of projectile motion passes through a horizontal line PQ which makes an angle of 37∫ with positive x-axis, xy plane is horizontal. The coordinates of the point where the particle will strike the line PQ is: (Take g = 10 m/s2

Answer 1

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)

 

Put A=45 deg; devide Y / X and equate it to tan37; we get value of time ‘t’;

Now, substitute the value of ‘t’ in eqn (1) to get X;

Similarly substitute the value of ‘t’ in eqn (2) to get Y;

Finally do not forget to add 2 to X;

You should get X+2 = 4.46 and Y = 1.857

Option A is closely matching;

Answer 2

Explanation:

Attachement :)))))

Thanks