Question

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

Answer:

The coordinates of point, Q are (10, 6,0).

Explanation:

Given a particle is projected from a point P (2, 0, 0)m

          velocity of a particle,  $u=10 m/s$

          angle with the horizontal. $\theta=45^{o}$

          angle of plane of projectile motion, $\theta=37^{o}$

         acceleration due to gravity, $g = 10 m/s^{2}$  

Range of projected particle,      

                  $R=\frac{u^{2} sin2\theta}{g}$

                      $=\frac{(10)^{2} sin2 \times 45}{10}=10m$

The coordinates of point, Q is

                 $\big (Rcos\theta+2, Rsin\theta, 0\big )=\big (10 \times cos37^{o} +2, 10 \times sin37^{o}, 0\big )$

                                                   $=\big (10 \times 0.8 +2, 10 \times 0.6, 0\big )$

                                                   $=\big (10, 6, 0\big )$

The coordinates of point, Q are (10, 6,0).