A body falling for 2 s covers a distance a which is equal to that covered in next l s. If `g = 10 ms^ -2`, the distance s is

A body falling for 2 seconds covers a distance S equal to that covered in next second. Taking g = 10m//s^(2),S =

A freely falling body, falling from a tower of heighth covers a distance h/2 in the last second of its motion The height of the tower is 10 ms^(-2) ):

A stone falls freely rest. The distance covered by it in the last second is equal to the distance covered by it in the first 2 s. The time taken by the stone to reach the ground is

A stone falls freely rest. The distance covered by it in the last second is equal to the distance covered by it in the first 2 s. The time taken by the stone to reach the ground is

A particle starts from the rest, moves with constant acceleration for 15s. If it covers s_1 distance in first 5s then distance s_2 in next 10s, then find the relation between s_1 & s_2 .

A particle starts from the rest, moves with constant acceleration for 15s. If it covers s_1 distance in first 5s then distance s_2 in next 10s, then find the relation between s_1 & s_2 .

A particle starts from the rest, moves with constant acceleration for 15s. If it covers s_1 distance in first 5s then distance s_2 in next 10s, then find the relation between s_1 & s_2 .

A body of mass (m) moving along a straight line covers half the distance with a speed of 2 m/s. The remaining half of the distance is covered in two equal tiem intervals with a speed of 3 ms^(-1) and 5 ms^(-2) respectively. The average speed of the particle for the entire journey is .