A man is standing on top of a building 100 m high. He throws two ball vertically, one at `t=0` and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. At `t=2`, both the balls reach to their and second ball is `+15m`.
Q. The speed of first ball is

Explanation

Let the speeds of the ratio two balls (1 and 2) be `v_1 and v_2 ` where if `v_1 =2v,v_2 =v.`
If ` y_1 and y_2 `are the distances covered by the balls 1 and 2 , respectively , before coming to rest then
` " " y_1 =(v_1^(2))/(2g) =(4v^(2))/(2g) and y_2 = (v_2^(2))/(2g) =(v^(2))/(2g) `
Since ` y_1 -y_2 15m ,( 4v^(2))/( 2g) -(v^(2))/( 2g) =15m or (3v^(2))/( 2g) =15 or v^(2) sqrt= ( 5mxx (2xx 10 m//s^(2)) or v= 10 m//s` Clearly ` " "v_1 =20 m//s and v_2 =10 m//s " "As " "y_1 =(v_1^(2))/(2g) =((20m)^(2))/(2xx10 m//s) =20 m ,`
`y_2 =y_1 -15m =5m`
If `t_2` is the time taken by the ball -2 to cover a distance of 5m , then from `y_2 =v_2 t-(1)/(2) gt_2^(2) ,5=10 t_2 -5t_2^(2)`
or ` " "t_2^(2) -2t_2+1=0 " ""hence" t_2 =1s`
Since `t_1` (time taken by ball -1 cover a distance of 20 m ) is 2s, time interval between the two throws ` =t_1-t_2 =2s -1s -1s`