Please Solve Q. D.2.... Answer given is Option (D)...

Please Solve Q. D.2.... Answer given is Option (D)... Projectile Motion D-2. A ball is horizontally projected with a speed v from the top of a plane inclined at an angle 450 with the horizontal. How far from the point of projection Will the ball strike the Plane? (B) 2v2 2üv2 (D) D-3. A particle is projected at angle 370 with the incline plane in upward direction with speed 10 m/s. The angleof incline plane is given 530. Then the maximum distance from the incline plane attained by the particle will be- 413m (B) 4m (C) 5m (D) zero DA. On an inclined plane of inclination 300, a ball is thrown at an angle of 600 with the horizontal from the foot of the incline with velocity of 10 a ms-l. If g = 10 ms-2, then the time in which ball will hit the inclined plane is - (A) 1 sec. (B) 6 sec. (C) 2 sec. (D) 4 sec. 1. ASSERTION / REASONING STATEMENT-I : In a projectile motion, the velocity at its highest point is zero. STATEMENT-2 : In a projectile motion from ground to ground projection, the acceleration is g downwards due to which speed of the projectile first decreases then increases to the same value. IA) Statement-I is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-I.

Dear student
Ans. to Q. D-2

For a horizontal projection of  a body with a velocity vh=x=12gt2x= horizontal range = vtThe plane is inclined at 45012gt2=vtt=2vgThereforex=v×2vg=2v2gwe have to find AB =xsin 45=22v2gRegards

  • 3
What are you looking for?