Two balls are dropped from different heights at different instants. Seconds ball is dropped 2 sec. after the first ball. If both balls reach the ground simultaneousl after 5 sec. of dropping the first ball. then the difference of initial heights of the two balls will be :- `(g=9.8m//s^(2))`

Options

A. 20 m
B. 80 m
C. 170 m
D. 40 m

Answer

B

Explanation

` -h_1 =-(1)/(2) g t_1^(2) ,-h_2 =-(1)/(2) g t_2^(2) t_1 =5 sec and t_2 =3 sec " "rArr " "h_1 -h_2 =(g)/(2) (5^(2) -3^(2) )= 80 m`