maximum height of trajectory|maximum distance travelled in projectile motion

Assertion : If time of flight in a projectile motion is made two times, its maximum height will become four times. Reason : In projectile motion H prop T^2 , where H is maximum height and T the time of flight.

Under what conditions the formulae of range, time of flight and maximum height can be applied directly in case of a projectile motion ?

A body is thrown up with a speed u , at an angle of projection theta If the speed of the projectile becomes u/sqrt2 on reaching the maximum height , then the maximum vertical height attained by the projectile is

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Statement-I : A projectile is thrown with an initial velocity of (ahat(i)+bhat(j)) m//s . If range of projectile is maximum then a=b . Statement-II : In projectile motion, angle of projection is equal to 45^(@) for maximum range condition.

The horizontal range is equal to two times maximum height of a projectile. The angle of projection ofthe projectile is:

A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance travelled by the body will be same in :-

Two projectile are projected which have the same range, if first projectile is projected at 30 degrees and its maximum height is h then the maxium height of other projectile is

Find the ratio of maximum horizontal range and the maximum height attained by the projectile. i.e. for theta_(0) = 45^(@)

In the previous problem , the maximum height attained by the ball and the distance traveled by the block , after collision will be