Question

A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff is 50.0 m and the angle of the cliff is θ = 20.0° 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 5.80 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

Answer 1

a)

initial velocity

v = d/t

v = 50 / 5.80

v = 8.62 m/s

kinematic equation

y = yo + vot + 1/2at^2

if water level is the origin with up as the positive direction

0 = 28 + (8.62sin20)t + 1/2(-9.81)t^2

0 = 28 - 2.9482t - 4.905t^2

4.905t^2 + 2.9482t - 28 = 0

solve quadratic equation

t = -.2.7085 s

or

t = 2.1075 s .................Ans

we ignore the negative time value as it occurred before the departure from the cliff.

b)

d = vt

d = (8.62cos20)(2.1075)

d = 17.07 m