Show that the horizontal range is maximum when the angle of projection is `45^(@)`

The range of a projectile for a given initial velocity is maximum when the angle of projection is 45^(@) . The range will be minimum, if the angle of projection is

Prove that the horizontal range id same when angle of projection is (i) greater than 45^(@) by certain value and (ii) less than 45^(@) by the same value.

Assertion: In projectile motion, when horizontal range is n times the maximum height, the angle of projection is given by tan theta=(4)/(n) Reason: In the case of horizontal projection the vertical increases with time.

Show that the horizontal range of a projectile is same for two angles of projection . Is the sum of maximum heights for these two angles dependent on the angle of projectile ?

Two balls are projected with the same velocity but with different angles with the horizontal. Their ranges are equal. If the angle of projection of one is 30° and its maximum height is h, then the maximum height of other will be

During a projectile motion if the maximum height equal the horizontal range, then the angle of projection with the horizontal is :

A body is thrown at an angle alpha to the horizontal with initial velocity v_0 .Assuming air drag to be negligible , find (i)the time of motion, (ii)maximum height of ascent and the horizontal range, (iii)the angle of projection at which they are equal , (iv)the angle of projection at which the horizontal range is maximum and the angle at which the vertical range is maximum.

Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is pi/3 and the maximum height reached by it is 102 m. Then the maximum height reached by the other in metres is

A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is

The angle of projection at which the horizontal range and maximum height of projectile are equal is