1.

A balloon is ascending vertically with an acceelration of `0.2 ms^(-2)` .Two stones are dropped from it at an interval fo `2 s`, the distance between then when the second stone dropped is (tanke g=9.i8 ma^(-2).

Answer» When the first stone is dropped from the balloon, let the initial velocity of the balloon be zero. Theis stone will fall under the efferct of gravity alone. The balloon is ascending with asceond stone placed in balloon after ` 2` second is:
` v=u+ at =0 + 0.2 xx 2 =0 0.4 ms^(-10`
Upward distanc ecoverd is
` S= ut + 1/2 at^2 =0 + 1/2 xx 0.2 xx 2^2 =0.4 m`
The second stone is dropped from ballon after `2 second`. The initial velocy of this stone is ` 0.4`ms^(-1)` up wards.
Taking vertcal downward motion of second stone fro time ` 1.5 second`, we have
`u =- 0.4 ms^(-1)` , a= 9.8 ms^(-2)` , S=S_1 , t =1.5 s`
`S_1 =- 0.4 xx 1.5 + 1/2 xx 9.8 xx (1.5)^@ = 10.425 m`
Taking vertical downward motion of first storne for time ``3.5 seconds` [= (2 + 1.5)` second] ,` we have
` u=0, a = 9.8 ms^(-2), S= S_2 , t =3.5 s`
S_2 =0 + 1/2 xx 9.8 xx (3.5 )^2 = 60 .025 m`
Separation between the two stones after time ` 1.5 cecond` when second stone is dropped
`= S_2- S_1 + S= 60.025 -10. 425 + 0.4 = 50 .0 m


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