A cliff diver positions herself on a cliff that angles downwards towards the edg

Question

A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff is 44.0 m and the angle of the cliff is ? = 18.0° below the horizontal. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 6.00 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 26.0 m before hitting the water

(a) After leaving the edge of the cliff, how much time does the diver take to get to the water in seconds?

(b) How far horizontally does the diver travel from the cliff face before hitting the water in meters ?

Explanation / Answer

Initial velocity

V = d/t = 44/6 = 7.33 m/sec

kinematic equation in y-direction

y = y0 + Voy*t + 0.5*a*t^2

Voy = -7.33*sin 18 deg = 2.265 m/sec

-ve sign because below the horizontal

0 = 26 - 2.265*t - 0.5*9.81*t^2

solvint this equation

t = 2.08 sec (A)

Now in horizontal direction

distance travelled will be

d = Vox*t

d = 2.08*7.33*cos 18 deg = 14.50 m (B)

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