For a given velocity of a projectile, the range is maximum when the angle of projection is

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

$${\text{y}} = {\text{x}}\tan \alpha + \frac{{{\text{g}}{{\text{x}}^2}}}{{2{{\text{u}}^2}{{\cos }^2}\alpha }}$$

For a given velocity of a projectile, the range is maximum when the angle of projection is

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