JEE MainPhysicsMotion in Two Dimensions
A projectile of mass m is fired from the ground. After t seconds, its inclination with the horizontal becomes zero. If the horizontal range of the projectile is R , its initial kinetic energy will be :
Options
- Am R^2 2 t^2 + m g^2 t^2 8
- Bm R^2 2 t^2 + 1 2 m g^2 t^2
- Cm R^2 8 t^2 + 1 2 m g^2 t^2
- Dm R^2 4 t^2 + m g^2 t^2
Correct answer
C. m R^2 8 t^2 + 1 2 m g^2 t^2
Step-by-step solution
When the inclination of the projectile with the horizontal becomes zero, its velocity is purely horizontal. This occurs at the maximum height of its trajectory. Therefore, the time taken to reach the maximum height is t . The vertical component of the initial velocity u_y can be found using the first equation of motion for the vertical direction ( v_y = u_y - gt ). At maximum height, v_y = 0 , so: 0 = u_y - gt u_y = gt The total time of flight of the projectile is twice the time to reach maximum height, so T = 2t .