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A stone is dropped from top of a tower 300 m high and at the same time another is projected vertically upward with a velocity of 100 ms^-1. Find when and where the two stones meet.

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3 s, 255.9m

5 s, 255.10m

2 s, 200.9m

4 s, 250.8m

answer is B.

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Detailed Solution

Time at which they meet is

$t = \frac{300}{100} = 3 s since t = \frac{h}{u_{relative}}$

Height at which they meet is

$s = ut - \frac{1}{2} {gt}^{2} = 100 \times 3 - \frac{1}{2} \times 9.8 \times 3^{2}$

s = h = 255. 9m

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