..A stone is thrown from 100m high building .....at the same time another stone is thrown at upward direction with a velocity 25m/s..(upar ki taraf feka gaya)......when and where the two stone meet...
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Let x and t be the displacement covered by the stone when it meet the another stone from upward and the time taken when the two stone meet
Then,
Displacement covered by the falling stone, h= ut +1/2gt²
= 0+1/2×9.8t²————(¹)
Again
Displacement covered by the upward stone,h = ut+1/2gt²
=› (100-x)=25t -1/2×9.8t²——(²)
From (¹) and(²)
100-4.9t²=25-4.9t²
=›t=25/100
=0.25
The required position of the ball above the ground =(100-4.9×0.25)m
=98.775m.
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