Two balloons are simultaneously released from two buildings A and B. Balloon from A rises with constant velocity 10 ms^-1, While the other one rises with constant velocity of 20 ms^-1 . Due to wind the balloons gather horizontal velocity V_x = 0.5 y, where y is the height from the point of release. The buildings are at a distance of 250 m and after some time t the balloons collide. Choose the correct option(s).

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difference in height of buildings is 50 m

t = 5 sec

difference in height of buildings is 100 m

t = 10 sec

answer is B, D.

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Detailed Solution

$(y_{B} - y_{A}) = 20 t - 10 t = 10 t$ ……..(i)

Difference in the height of building

$(X_{B} - X_{A}) = \int_{0}^{t} (0 .5 y_{B}) dt - \int_{0}^{t} (0 .5 y_{A}) dt$ …….(ii)

Separation between the buildings = 250 m

Hence $\int_{0}^{t} 0 .5 \times 20 t . dt - \int_{0}^{t} 0 .5 \times 10 t . dt = 250$

$5 \int_{0}^{t} t . dt = 250$

Hence $5 \frac{t^{2}}{2} = 250 \Rightarrow t = 10 s$

Difference in the height of building = 10t $= 10 \times 10 = 100 m$

Difference in heights $= y_{2} - y_{1} = 20 t - 10 t = 100 m$

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