A diver runs horizontally off the end of a 3 m high diving board at 6 m/s. a.) H

Question

A diver runs horizontally off the end of a 3 m high diving board at 6 m/s.

a.) How long does it take for the diver to hit the water?

b.) How far horizontally does the diver end up from the board?

c.) What is the size of the horizontal (x) component of the diver?s velocity just before he hits the water

d.) What is the size of the vertical (y) component of the diver's velocity just before he hits the water?

e.) What is the speed of the diver just before he hits the water?

f.) If the diver instead horizontally ran off the end at 12 m/s, which of the answers for a-e would change?
Select all that apply to get credit.

Explanation / Answer

a. The diver has no vertical velocity component when he runs of the board. Thus vy_initial = 0. Use y = vy_initial*t + (1/2)at^2. ==> -3 = 0*t - 4.9t^2 ==> t = 0.782 s. b. The diver's horizontal velocity component remains constant since there is no acceleration in the horizontal direction. Use t from part a. Use x = vx*t = (6 m/s)(0.782 s) = 4.69 m. c. Like I mentioned in part b, the horizontal component of the diver's velocity remains constant. The velocity is 6 m/s. d. Use vy_final = vy_initial + at, where a = -9.8 and vy_initial = 0. Use t from part a. vy_final = 0 - 9.8 m/s^2 * 0.782 s = -7.66 m/s. e. The speed of the diver is the magnitude of his velocity vector. So speed = sqrt(vx^2 + vy_final^2) = sqrt(6^2 + (-7.66)^2) = 9.73 m/s. (I may have accrued some rounding error here, but the answer should be close enough).

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