Related Questions

The range of projectile will be maximum for a given velocity of projectile, when the angle of projection ($$\alpha $$) is
A $$\frac{\beta }{2}$$
B 30° + $$\frac{\beta }{2}$$
C 45° + $$\frac{\beta }{2}$$
D 60° + $$\frac{\beta }{2}$$
For the given values of initial velocity of projection and angle of inclination of the plane, the maximum range for a projectile projected upwards will be obtained, if the angle of projection is
A $$\alpha = \frac{\pi }{4} - \frac{\beta }{2}$$
B $$\alpha = \frac{\pi }{2} + \frac{\beta }{2}$$
C $$\alpha = \frac{\beta }{2} - \frac{\pi }{2}$$
D $$\alpha = \frac{\pi }{4} - \frac{\beta }{4}$$
E $$\alpha = \frac{\pi }{2} - \frac{\beta }{2}$$
The range of projectile on a downward inclined plane is ________ the range on upward inclined plane for the same velocity of projection and angle of projection.
A Less than
B More than
C Equal to
D None of these
The range of a projectile is maximum, when the angle of projection is
A 30°
B 45°
C 60°
D 90°
For maximum range of a projectile, the angle of projection should be
A 30°
B 45°
C 60°
D None of these
The angle of projection at which the horizontal range and maximum height of a projectile are equal to
A 36°
B 45°
C 56°
D 76°
The angle of projection for a range is equal to the distance through which the particle would have fallen in order to acquire a velocity equal to the velocity of projection, will be
A [[[
If the direction of projection bisects the angle between the vertical and the inclined plane, then the range of projectile on the inclined plane is
A Zero
B Maximum
C Minimum
D None of these
The direction of projection should bisect the angle between the inclined plane and the vertical for a range of a projectile on inclined plane
A To be zero
B To be maximum
C To be minimum
D None of these
The Cartesian equation of trajectory is (where u = Velocity of projection, $$\alpha $$ = Angle of projection and x, y = Co-ordinates of any point on the trajectory after t seconds.)
A $${\text{y}} = \frac{{{\text{g}}{{\text{x}}^2}}}{{2{{\text{u}}^2}{{\cos }^2}\alpha }} + {\text{x}}\tan \alpha $$
B $${\text{y}} = \frac{{{\text{g}}{{\text{x}}^2}}}{{2{{\text{u}}^2}{{\cos }^2}\alpha }} - {\text{x}}\tan \alpha $$
C $${\text{y}} = {\text{x}}\tan \alpha - \frac{{{\text{g}}{{\text{x}}^2}}}{{2{{\text{u}}^2}{{\cos }^2}\alpha }}$$
D $${\text{y}} = {\text{x}}\tan \alpha + \frac{{{\text{g}}{{\text{x}}^2}}}{{2{{\text{u}}^2}{{\cos }^2}\alpha }}$$

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