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A stone is allowed to fall from the top of the tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. The two stones will meet above the ground at height of
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Suppose the two stones meet at a height h from the ground. The stone that is dropped from above will cover (100 - h) before meeting the stone that is thrown from ground. Let, time taken is t. So, 100−h=(0)t+12(10)t2 Or 100−h=5t2 ?(i) Similarly, for stone thrown from ground, h=25t−12(10)t2 Or h=25t=−5t2 ?(ii) Adding (i) and (ii), 100 = 25t Or t=4 s Using (ii), h=25×4−5(4)2=100−80=20m
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Question Text | A stone is allowed to fall from the top of the tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. The two stones will meet above the ground at height of |
Answer Type | Text solution:1 |
Upvotes | 150 |