a pendulum swinging back and forth rises at the end of its swing to position 20 cm higher than its lowest point.how fast is it going at the lowestpoint.

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Solution

Dear Student ,
Here in this case maximum height is 20 cm = 0.2 m
Now from the conservation of energy we can write that ,
$m g h = \frac{1}{2} m v^{2} \Rightarrow g h = \frac{1}{2} v^{2} \Rightarrow 9 \cdot 8 \times 0 \cdot 2 \times 2 = v^{2} \Rightarrow v = 1 \cdot 98 m / s$
Regards

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a pendulum swinging back and forth rises at the end of its swing to position 20 cm higher than its lowest point.how fast is it going at the lowestpoint.

Dear Student , Here in this case maximum height is 20 cm = 0.2 m Now from the conservation of energy we can write that , mgh=12mv2⇒gh=12v2⇒9·8×0·2×2=v2⇒v=1·98 m/s Regards

Dear Student ,
Here in this case maximum height is 20 cm = 0.2 m
Now from the conservation of energy we can write that ,
$m g h = \frac{1}{2} m v^{2} \Rightarrow g h = \frac{1}{2} v^{2} \Rightarrow 9 \cdot 8 \times 0 \cdot 2 \times 2 = v^{2} \Rightarrow v = 1 \cdot 98 m / s$
Regards