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A 822 N diver drops from a board 12.0 m above the water’s surface. Find the diver’s speed 5.00 m above the water’s surface. ------------ m/s. What is the diver’s speed just before striking the water? |
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Answer» Using conservative of energy mgh1 + 1/2 m(v1)2 = mgh2 + 1/2 m(v2)2 [9 x 12 + 1/2 (o)2] = [9.8 x 5 + 1/2 (v2)2] \(\because\) g = 9.8 [9.8 x 12 + 0] = [49 + 1/2 + \(\text v_2^2\)] 117.6 = 46 + 1/2 \(\mathrm v_2^2\) 1/2 \(\text v_2^2\) = 117.6 - 49 \(\text v_2^2\) = 68.6 x 2 \(\text v_2^2\) = 137.2 ⇒ 11.71 m/s before srinking the water mgh1 + 1/2 m\(\text v_1^2\) = mgh2 + 1/2 m\(\text v_1^2\) [9.8 x 12 + 1/2 x [0]2] = [9.8 x 0 + 1/2 + \(\text v_2^2\)] 117.6 = 1/2 x \(\text v_2^2\) \(\text v_2^2\) = 117.6 x 2 \(\text v_2^2\) = 235.2 v2 = \(\sqrt{235.2}\) v2 = 15.33 m/s |
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