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 arrows will be at the same height above the ground at t = 12.5 s
B. The two arrows will reach back their starting points at t = 20 s and at t = 25 s
C. The ratio of the speeds of the first and second arrows at t = 20 s will be 2 :
D. The maximum height attained by either arrow will be 1000 m

Answer

A, B, C

Explanation

Let they meet at height after time t
` h= 100 t -(1)/(2) g t^(2) (t-5) ^(2) to ` for second arrow `" "rArr t=12.5 s ` (after (A) is correct)
Time of flight of first arrow` T=( 2u)/(g) =(2xx 100 )/( 10 ) =20s `
Second arrow will reach after 5 s of reaching first. So (B) is correct.
At t =20 s , ` v_1=100 -10 xx 20 =-100 ms^(-1) , v_2 =100 -10 xx 15 =-50 ms^(-1) `
Ratio: `(v_1)/(v_2) =2:1 ` So (C) is correct
Maximum height attained ` , " "H=(u^(2))/(2g) =((100)^(2))/(2xx 10 ) =50 m. `" "`Hence (D) is