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Q.

From the top of the tower of height 400 m, a 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 ground 400−s=12gt2.........(i)and for second ball: s=50t−12gt2Adding 50t=400, we gett= 8 sec.Now substituting the value of ‘t’ in (i), we get s=50×8−12×10×64=80m

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