Two balls are dropped from the same point after an interval of 1s. If acceleration due to gravity is `10 m//s^(2)`, what will be their separation 3 seconds after the release of first ball?

Two balls are released from the same height at an interval of 2s. When will the separation between the balls be 20m after the first ball is released? Take g=10m//s^(2) .

A ball is projected from a point O as shown in figure. It will strike the ground after (g = 10 m//s^(2))

A ball 4 s after the instant it was thrown from the ground passes through a point P, and strikes the ground after 5 s from the instant it passes through the point P. Assuming acceleration due to gravity to be 9.8 m//s^(2) find height of the point P above the ground.

A ball 4 s after the instant it was thrown from the ground passes through a point P, and strikes the ground after 5 s from the instant it passes through the point P. Assuming acceleration due to gravity to be 9.8 m//s^(2) find height of the point P above the ground.

S = ut-1/2 g t^2 is a formulat which given the distance S in meters travelled by a ball from the thrower's hadns if it is thrown upwards with an initial velocvity of u m//s after a time of t seconds. G is the acceleration due to the gravity and is 9.8 m//s^2 If a ball is thrown upwards at 14m//s how high has it gone after 1 second?

Two balls are dropped simultaneously from two points separated by a vertical height of 6 m. The distance of separation between them after next 2s is

Two balls are dropped simultaneously from two points separated by a vertical height of 6 m. The distance of separation between them after next 2s is