A high diver jumps vertically upward at 5 m/s. The diving board is 10 m above the water. How long after jumping does he hit the water? Use 10 m/s-squared for g.
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(2 x 5)/10 = 1 secs
v^2 = u^2 + 2gs = 5^2 + (2 x 10 x 10) = 225
sq-root(225) = v = 15 m/s
10/((5 + 15)/2) = t = 1 sec
total = 2 secs
v^2 = u^2 + 2gs = 5^2 + (2 x 10 x 10) = 225
sq-root(225) = v = 15 m/s
10/((5 + 15)/2) = t = 1 sec
total = 2 secs
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Time going up = t = initial velocity/ g
Height going up = h = g/2 x t^2
Distance going down = d = (10 + h)
Time going down = T = square root of (2d/g)
Total time = t + T
Height going up = h = g/2 x t^2
Distance going down = d = (10 + h)
Time going down = T = square root of (2d/g)
Total time = t + T
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50