The velocity of a freely falling body changes as `g^ph^q`where g is acceleration due to gravity and h is the height. The values of p and q are

Let g be the acceleration due to gravity on the earth's surface.

The velocity of a freely falling body is a function of the distance fallen through (h) and acceleration due to gravity g . Show by the method of dimensions that v = Ksqrt(gh) .

The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. The velocity of the body as it reaches the ground is

The value of the acceleration due to gravity g on earth depends upon

Find ratio of acceleration due to gravity g at depth d and at height h where d=2h

Acceleration the due to gravity become [(g)/(2)] at a height equal to

Let us consider an equation (1)/(2)muepsilon^2 = mgh, Where m is the mass of the body, upsilon its velocity, g is acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct.

Find the value of a and b in the following cases : (a) The velocity v of the ball falling freely under gravity is proportional to g^(a) h^(b) , where g is the acceleration due to gravity, h is the height from which the ball is dropped. (b) The kinetic energy K of a rotating body is proportional to I^(a) omega^(b) where I is the moment if inertia and omega is the angular speed. (c ) The time-period T of spring pendulum is proportiona to m^(a) k^(b) , where m is the mass of block attached to the spring and k is the spring constant. The speed of sound v in a gaseous medium is proportional to P^(a) rho^(b) , where P is the pressure and rho is the density of medium.

What is acceleration due to gravity? Obtain relations to show how the value of 'g' changes with Altitude.

What is acceleration due to gravity? Obtain relations to show how the value of 'g' changes with Depth.