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 48.0 m and the angle of the cliff is = 21.0° 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 = 30.0 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?

What is the speed of the diver as she leaves the cliff? In what direction is she traveling? What constant acceleration relationship will allow you to find the time with the given information? s

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

What is the initial position and final position of the diver as she leaves the cliff? Can you use this to find the diver's time of flight? Ignoring air resistance, is there any acceleration in the x (horizontal) direction? m

Explanation / Answer

a. apply time of descent t = 2h/g

t = 2*30/9.8

t = 6.12 secs

here speed u = x/t = 48/5.5 = 8.72 m/s

g is accleration due to gravity = 9.8 m/s

Speed u = x/t = 48/5.5 = 8.72 m/s

Downward direction

a = g = accleration = 9.8m/s^2

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B. Range R = u^2 sin 2 theta/g

R = (8.72 * 8.72 * sin (2 * 21)/9.81

R   = 5.18 m

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time of flight t = 2u/g

t = 2* 8.72/9.81

t = 1.77s ecs

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yes, a = v-u/t

a = 8.72/1.77

a = 4.9 m/s^2

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