A ball is dropped from a height ,if it takes 0.2 second to cross the last 6 m before hitting the ground. Find the height from which it was dropped. (Take g = 10 m/sec)
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Sol. For last 6 m distance travelled s = 6 m, u = ? t = 0.2 sec, a = g = 9.8 m/s2 S = ut + ½ at2 ⇒ 6 = u (0.2) + 4.9 x 0.04 ⇒ u = 5.8/0.2 = 29 m/s. For distance x, u = 0, v = 29 m/s, a = g = 9.8 m/s2 S = (V^2-u^2)/2a = (〖29〗^2 - 0^2 )/(2 x 9.8 ) = 42.05 m Total distance = 42.05 + 6 = 48.05 = 48 m.
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Explanation:
Sol. For last 6 m distance travelled s = 6 m, u = ? t = 0.2 sec, a = g = 9.8 m/s2 S = ut + ½ at2 ⇒ 6 = u (0.2) + 4.9 x 0.04 ⇒ u = 5.8/0.2 = 29 m/s. For distance x, u = 0, v = 29 m/s, a = g = 9.8 m/s2 S = (V^2-u^2)/2a = (〖29〗^2 - 0^2 )/(2 x 9.8 ) = 42.05 m Total distance = 42.05 + 6 = 48.05 = 48 m.
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