Question

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.74 m above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 2.05 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

Answer 1

Since diver run horizontally off the cliff, So initial vertical speed of diver is zero

Now Using 2nd kinematic equation in vertical direction

h = V0y*t + (1/2)*a*t^2

h = height of cliff = 8.74 m

V0y = 0 m/sec

a = acceleration in vertical direction = g = 9.81 m/sec^2

So,

8.74 = 0*t + (1/2)*9.81*t^2

t = sqrt (2*8.74/9.81)

t = 1.335 sec

Now if average angular speed of diver is 2.05 rev/sec, then

Number of revolutions = = w*t

= 2.05*1.335

= 2.74 rev

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