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. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform, she starts flipping, spinning with a constant angular acceleration of 15.0rad/s2. How many revolutions does she make before hitting the water? (Note that maintaining constant angular acceleration in the air is not realistic, but let’s do it here anyway.)\

User Saqueib
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1 Answer

3 votes

Answer:

2.39 revolutions

Step-by-step explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is


h = gt^2/2


10 = 10t^2/2


t^2 = 2


t = √(2) = 1.414 s

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:


\theta = \alpha t^2/2


\theta = 15 * 2/2 = 15 rad

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

User Chris Vietor
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