If a particle takes `t` second less and acquire a velocity of `vms^(-1)` more in falling through the same disance on two planets where the accelerations due to gravity are `2g` and `8g` respectively, then `v=xgt`. Find value of `x`

A particle takes n seconds less and acquires a velocity u m/sec. higher at one place than at another place in falling through the same distance. Calculate the product of the acceleration due to gravity at these two places.

A particle takes n seconds less and acquires a velocity u m/sec. higher at one place than at another place in falling through the same distance. Calculate the product of the acceleration due to gravity at these two places.

Average densities of two planets is same but their radii are R_1 and R_2 . If the acceleration of gravity on planets are g_1 and g_2 respectively, then

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

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

A freely body acquires a velocity of g^(x)h^(y) after falling through a height h. Using dimensions find the values fo x and y.

Two planets have the same average density but their radii are R_(1) and R_(2) . If acceleration due to gravity on these planets be g_(1) and g_(2) respectively, then

Two particles are projected vertically upward with the same velocity on two diferent planes with accelerations due to gravities g_(1) and g_(2) respectively. If they fall back to their initial points of projection after lapse of time t_(1) and t_(2) respectively. Then