An object is thrown vertically upward with a speed `u_(1)` and it travels 8 m in the last secon of its upward motion. If the object is thrown upward with a speed `u_(2)` which is twice of `u_(1)`, the distance now travelled by the object during the last second of its upward journey will be:

Options

A. 32 m
B. 16 m
C. 12 m
D. 8 m

Answer

D

Explanation

The distance travelled in last second is always equal because it is equal to
`s = (1)/(2) xx g (t)^(2)`