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Q.

From the top of the tower of height 400 m, & ball is dropped by a man, simultaneously from the base of the tower, another ball is thrown up with a velocity 50 m/s; at whatdistance will they meet from the base of the tower?

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a

100 m

b

320 m

c

80 m

d

240 m

answer is C.

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Detailed Solution

Let the first ball meet at a height s from ground400−s=12gt2 ……..(i) and for second ball: s=50t−12gt2 ……..(ii) Adding 50t=400, we get t=8sec Now substituting the value of ' t ' in (ii), we get s=50×8−12×10×64=80m

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