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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 what distance will they meet from the base of the tower?

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320 m

240 m

100 m

80 m

answer is C.

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Detailed Solution

$400 - s = \frac{1}{2} g t^{2} ... (i)$

$a n d s = 50 t - \frac{1}{2} g t^{2} .. (i i)$

Adding, 50t = 400

Or t = 8 s.

$s = 50 \times 8 - \frac{1}{2} \times 10 \times 64 = 80 m$

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