What is the ball’s speed at the lowest point of its trajectory?

A pendulum is made by tying a 530 g ball to a 52.0 cm-long string. The pendulum is pulled 21.0-degree to one side, then released.

Answer 1

Use the law of conservation of energy.

(1/2)mV^2 = mgL(1 – cos 21)

where

m = mass of the ball = 530 g (given)
V = speed of the ball at its lowest point
L = length of string = 52 cm. (given)
g = acceleration due to gravity = 980 cm/sec^2 (constant)

Since “m” appears on both sides of the equation, it will cancel out and the above simplifies to

(1/2)V^2 = gL(1 – cos 21)

Substituting appropriate values,

(1/2)(V^2) = 980(52)(1 – 0.9334)

V = 82.28 cm/sec.

Hope this helps.

Answer 2

Using trig, first calculate the ball’s vertical displacement at its highest point, i.e. when the string is at a 21 degree angle to the vertical. You will find that it is 104*Sin^2(10.5) = 3.4538 cm. above its lowest point.

Now calculate the speed of the ball after falling this height under a gravitational acceleration of 981 cm/sec^2 using v = (2ah)^0.5

The answer is 82.32 cm./sec

Answer 3

the question is too hard

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