(i) When the stone from the top of the tower is thrown, Initial velocity u = 0 Distance travelled = x Time taken = t Therefore, Height, h = (100 – x) Initial velocity, u = 0 g = 10 m/s2 (As the stone is falling down) h= ut+1/2gt2 Substituting the values in above equation (100 – x) = 5t2 (i) Height, h = x Time, t =? We know that, s= ut+1/2gt2 (100 – S) = 25t + 1/2 (-10) t² = 25t – 5t² On adding the above equations We get 100 = 25t or t = 4 s After 4sec, two stones will meet From (a) x = 5t2 = 5 x 4 x 4 = 80m. Putting the value of x in (100-x) = (100-80) = 20m. This indicates that after 4seconds, 2 stones meet a distance of 20 m from the ground. (i) When the stone from the top of the tower is thrown, Initial velocity u = 0 Distance travelled = x Time taken = t Therefore, Height, h = (100 – x) Initial velocity, u = 0 g = 10 m/s2 (As the stone is falling down) h= ut+1/2gt2 Substituting the values in above equation (100 – x) = 5t2 (i) Height, h = x Time, t =? We know that, s= ut+1/2gt2 (100 – S) = 25t + 1/2 (-10) t² = 25t – 5t² On adding the above equations We get 100 = 25t or t = 4 s After 4sec, two stones will meet From (a) x = 5t2 = 5 x 4 x 4 = 80m. Putting the value of x in (100-x) = (100-80) = 20m. This indicates that after 4seconds, 2 stones meet a distance of 20 m from the ground.
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