If a freely falling body travels in the last second, a distance equal to the distance travelled by

A freely falling body travels-- of total distance in 5th second

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

Distance travelled by a freely falling body in the nth second is equal to 'n' sec find the value of n

A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 seconds. It remained in air for :-

A freely falling body travelled x m in n^("th") second. Then distance travelled in n-1^("th") second is

A stone falls freely from rest and the total distance covered by it in last second of its motion equals the distance covered by it in the first 3s of its motion. How long the stone will remain in the air?

A freely falling body takes t second to travel (1//x)^(th) distance then time of descent is

A freely falling body takes 't' second to travel first 1/xth distance.Then,time of descent is

A freely falling body takes T second to travel first (1/x)th distance.Then,time of descent is

Assertion : Distance travelled by the particle in the nth second has dimensios of length. Reason :It is the distance travelled by the partical in the given time.