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anonymous
 4 years ago
I solid iron ball of mass 770kg is used on a building site. The ball is suspended by a rope from a crane. The distance from the point of suspension to the centre of the mass of the ball is 12 m.The ball is pulled back from the vertical and then realeased, it falls through a vertical height of 1.6m and strikes the wall: What is the speed of the ball just before impact?
anonymous
 4 years ago
I solid iron ball of mass 770kg is used on a building site. The ball is suspended by a rope from a crane. The distance from the point of suspension to the centre of the mass of the ball is 12 m.The ball is pulled back from the vertical and then realeased, it falls through a vertical height of 1.6m and strikes the wall: What is the speed of the ball just before impact?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Don't exactly know what the means, ball is pulled back from the vertical then released.. Not worded.. Are you saying it was dropped 1.6m? If so then it's giving you a lot of information that you don't even need. you just need conservation of energy or kinematics to solve the problem. 1/2*m*v^2 = mgh mass cancels, solve for v.  Otherwise, you will need to restart the problem or be more clear.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually Made a mistake on that.. Going to do it again. From a balistics pendulum as an example, which is probably where you are at... Triangles, L = 12m, x_0 = 1.6, we need to find the angle (theta) sin (theta) = 1.6/12 = 0.1333.... theta = ArcSin(0.1333...) = 0.1337 rad Then, this is the height that the mass vertically raises in it's arc y_2 = LL*cos(theta) = 0.107 m use y_2 in a kinematic swing... \[v=\sqrt{2gy_2}=1.45 m/s\]
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