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 θ = 19.0° 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 6.30 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 22.0 m before hitting the water. How far horizontally does the diver travel from the cliff face before hitting the water?

Answer 1

Speed that the diver leaving the cliff

u = Speed that the diver runs towards the edge

u = (48 m)/(6.3 s)

u = 7.62 m/s

Initial speed = u = 7.62*sin19^o m/s

a = 9.81 m/s²

s = 22 m

s = u*t + ½*a*t²

22 = (7.62*sin19)*t + 0.5*(9.81)*t²

22 = 2.48*t + 4.905*t²

t = 1.88 s or t = -2.38 s

Time taken = 1.88s

Horizontal distance = (7.62 cos19 m/s)*1.88s

D = 13.54 m