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A stone is thrown vertically upwards from the ground with a velocity 15 m/s. At the same instant a ball is dropped from a point directly above the stone from a height of 30 m. At what height from the ground will the stone and the ball meet and after how much time? (Use g = 10 m/s2 for ease of calculation). |
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Answer» Let the stone and the ball meet after time t0. From second equation of motion, the distances travelled by the stone and the ball in that time is given as, Sstone = 15 t0 – \(\frac{1}{2}\) gt02 Sball = \(\frac{1}{2}\) gt02 When they meet. Sstone + Sball = 30 ∴ 15t0 – \(\frac{1}{2}\) gt02 + \(\frac{1}{2}\) gt02 = 30 t0 = \(\frac{30}{15}\) = 2 s ∴ Sstone = 15 (2) – \(\frac{1}{2}\) (10) (2)2 = 30 – 20 = 10 m The stone and the ball meet at a height of 10 m after time 2s. |
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