A girl is swinging on a swing set 2 m above the ground. The length of the swing

Question

A girl is swinging on a swing set 2 m above the ground. The length of the swing is 4 m. When the girl hits 45 degrees, she jumps, and lands 8 meters away. Find the centripetal acceleration of the swing, the velocity of the girl when she leaves the swing, and the max height the girl reaches after she jumps. I was able to find the height of the girl when she jumps by creating a triangle. Since this is the distance below the swing set plus 2 m: 4-2.82 + 2 = 3.19 m However. I don't know where to go from hero. I know that I need the velocity of the girl when she leaves the swing to got the centripetal acceleration. How do I get this? Is 45 the angle at which she leaves the swing? Even if it is, how do I get the velocity with the theta when the ground isn't level?

Explanation / Answer

girls velocity at point of jump is 45 degree to the horizontal. Acceleration due to gravity acts only on verticle component of velocity. so v(horizontal)*t = 8 also you have the height =3.19 m and you can make another equation verticle motion -3.19 = v(verticle)*t - 1/2 *g*t^2 but v(horizontal)=v(verticle) putting these values you can get horizontal/verticle component of velocity and hence the total velocity of the girl you can derive the acceleration by v^2/r relation and if you have v you can easily calculate the max height obtained

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