Question

A body is dropped from a certain height. 02 seconds later another body is projected vertically upward from the same point at 20m/s. The distance between the first and second bodies 5 seconds after the projection of the second body is.........​m.
(acceleration due to gravity = 10 m/s2)

Answer 1

Explanation:

1. dropped body u=0
2. s=ut+1/2at²
3. s=0+1/2(10)(5)²
4. s=125m
5. projected body u=20
6. s at 5 seconds
7. s=20(5)-10(25)
8. s=-150
9. so difference between their distance is 125-(-150)
10. =275