Horizontal Range Of The Escaping Liquid

In previous problem, the horizontal range of the stone is:

If R is the horizontal range for theta inclination and h is the maximum height reached by the projectile, Then maximum range is

An object is thrown along a direction inclined at an angle of 45^(@) with the horizontal direction. The horizontal range of the particle is equal to

Assertion: The maximum horizontal range of projectile is proportional to square of velocity. Reason: The maximum horizontal range of projectile is equal to maximum height attained by projectile.

The maximum height attained by a projectile when thrown at an angle theta with the horizontal is found to be half the horizontal range. Then theta is equal to

The maximum height attained by a projectile when thrown at an angle theta with the horizontal is found to be half the horizontal range. Then theta is equal to

From the ground, a projectile is fired at an angle of 60 degree to the horizontal with a speed of 20" m s"^(-1) The horizontal range of the projectile is

The horizontal range for projectile is given by