At `t=0`, an arrow is fired vertically upwards with a speed of `100 m s^(-1)`. A second arrow is fired vertically upwards with the same speed at `t=5 s`. Then .

Options

A. The two bullets will be at the same height above the ground at `t=12.5 s`
B. The two bullets will reach back their starting point at `t=20s` and at `t=25s`
C. The ratio fo the speeds of the first and second bulletsat `t=20s` will be `2:1`
D. The maximum height attained by either bullet will be 1000 m

Answer

A, B, C

Explanation

Let they meet at height `h` after time t.
`h=100t-(1)/(2)g t^(2)to` for first bullet
`=100(t-5)-(1)/(5)g(t-5)^(2)to` for second bullet
`impliest=12.5sec` (after solving) so (a) is correct.
`T=(2u)/(g)=(2xx100)/(10)=20s`
Second bullet will reach after 5 s of reaching first So (b) is correct.
`v_(1)=100-10xx20=-100ms^(-1)`
`v_(2)=100-10xx15=-50ms^(-1)`
ratio: `(V_(1))/(V_(2))=2:1` so (C) is correct maximum height attained
`H=(u^(2))/(2g)=((100)^(2))/(2xx10)=500m`, hence (d) is incorrect.