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 47.0 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 5.50 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 28.0 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? s (b) How far horizontally does the diver travel from the cliff face before hitting the water? m

Explanation / Answer

Velocity of the driver at the edge of the cliff, v = distance/time
= 47/5.5
= 8.55 m/s

Initial vertical velocity = v sin(18)
= 8.55 x sin(18)
= 2.64 m/s
Initial horizontal velocity = v cos(18)
= 8.13 m/s.

a)
Consider the vertical motion of the diver.
Initial vertical velocity, u = 2.64 m/s
Acceleration, g = 9.81 m/s2.
Distance traveled = Height of the cliff = 28 m.
Using the formula, h = ut + 1/2 gt2,
28 = 2.64 t + 4.905 t2.
Solving this quadratic equation,
t = 2.135 and - 2.6735
Since time is positive, t = 2.135 s.

b)
Horizontal velocity = 8.13 m/s.
Time taken, t = 2.135 s
Since there is no horizontal acceleration,
Distance traveled, D = velocity x time
= 8.13 x 2.135
= 17.35 m.

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