A fort is at the top of a hill of height h above sea level .The greatest horizontal distance from which a gun in the ship can hit the fort. if the muzzle velocity is `sqrt(2gk)` will be (g is acceleration due to gravity, k is a positive constant)

T=(R^(2))/(r^(2))sqrt((2h)/(g)) Suppose an open cylindrical tank has a round drain with radius t in the bottom of the tank. When the tank is filled with water to a depth of h centimeters, the time it takes for all the water to drain from the tank is given by the formula above, where R is the radius of the tank (in centimeters) and g=980"cm/"s^(2) is the acceleration due to gravity. Suppose such a tank has a radius of 2 meters and is filled to a depth of 4 meters. About how many minutes does it take to empty the tank if the drain has a radius of 5 centimeters? (1 meter=100 centimeter).

A projectile is fired horizontally with velocity of 98 m/s from the top of a hill 490 m high. Find (a) the time taken by the projectile to reach the ground, (b) the distance of the point where the particle hits the ground from foot of the hill and (c) the velocity with which the projectile hits the ground. (g= 9.8 m//s^2) .

A body is projected up a smooth inclined plane of length 20sqrt(2)m from point A as shown in the figure. The top of B of the inclined plane is connected to a well of diameter 40 m. If the body just manages to cross the well then the velocity of projection is (Acceleration due to gravity, g=10ms^(-2) )

An arrow is launched upward with an initial speed of 70m/s ( meters per second). The equation v^(2)=v_(0)^(2)-2gh describes the motion of the arrow, where v_(0) is the initial speed of the arrow, v is the speed of the arrow as it is moving up in the air, h is the height of the arrow above the ground, t is the time elapsed since the arrow was projected upward, and g is the acceleration due to gravity ( approximately 9.8 m//s^(2) ). What is the maximum height from the ground the arrow will rise to the nearest meter?

A ball is thrown from a point in level with velocity u and at a horizontal distance r from the top os a tower of height h . At what horizontal distance x from the foot of the tower does the ball hit the ground ?

A body is thrown from the surface of the earth with velocity u m/s. The maximum height in metre above the surface of the earth upto which it will reach is (where, R = radish of earth, g=acceleration due to gravity)

A body of mass m thrown horizontally with velocity v, from the top of tower of height h touches the level ground at a distance of 250m from the foot of the tower. A body of mass 2m thrown horizontally with velocity v//2 , from the top of tower of height 4h will touch the level ground at a distance x from the foot of tower. The value of x is

A body of mass m thrown horizontally with velocity v, from the top of tower of height h touches the level ground at a distance of 250m from the foot of the tower. A body of mass 2m thrown horizontally with velocity v//2 , from the top of tower of height 4h will touch the level ground at a distance x from the foot of tower. The value of x is

A particle is ejected from the tube at A with a velocity V at an angle theta with the verticle y-axis at a height h above the ground as shown. A strong horizontal wind gives the particle a constant horizontal acceleration a in the positive x-direction. If the particle strikes the ground at a point directly under its released position and the downward y-acceleration is taken as g then find h.

A hemi-spherical tank of radius 2 m is initially full of water and has an outlet of 12c m^2 cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law v(t)=0.6sqrt(2gh(t)), where v(t) and h(t) are, respectively, the velocity of the flow through the outlet and the height of water level above the outlet and the height of water level above the outlet at time t , and g is the acceleration due to gravity. Find the time it takes to empty the tank.