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A stone of mass ‘m’ is to be thrown to a height h1. What is the acceleration of the stone?2. With what minimum velocity should it be thrown.3. At what height does the KE and PE become equal?4. Find the velocity at that height |
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Answer» 1. –g or –9.8m/s. 2. v = 0, a = -g, S = h Substitute this values in V2 = u2 + 2as we get 0 = u2 – 2gh u = \(\sqrt{2gh}\) 3. at \(\frac{h}{2}\)., KE and PE are equal. 4. V2 = U2 + 2aS = U2 – 2g \(\frac{h}{2}\) (u2 = 2gh) = U2 – \(\frac{U^2}{2}\), V2 = \(\frac{U^2}{2}\), V = \(\frac{U}{√2}\). |
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