How to change Draggable component’s original position

I’m working with the react-native-draggable component.

return (
    <View >
        <Draggable x={75} y={100} renderSize={56} renderColor='black' renderText='A' isCircle shouldReverse onShortPressRelease={()=>alert('touched!!')}/> 
        <Draggable x={200} y={300} renderColor='red' renderText='B'/>
        <Draggable/>
    <Draggable x={50} y={50}>
        <YourComponent/>
    </Draggable>
    </View>
);

By setting the shouldReverse prop to true, the draggable item smoothly returns to its original position.

However, I can’t seem to dynamically update the “original position” of the draggable item by updating the position.x and position.y values, as in the following code:

    <View style={styles.answerContainer}>
        {answerParts.map((item, index) => (
            <Draggable
                key={item.index}
                x={item.position.x}
                y={item.position.y}
                disabled={item.placed}
                shouldReverse={!item.placed}
                onDragRelease={(event) => onReleaseItem(item, event)}
            >
                <View style={styles.draggableItem}>
                    <Text style={styles.draggableText}>{item.text}</Text>
                </View>
            </Draggable>
        ))}
    </View>

Any thoughts on how I can properly change the original location of a draggable React Native component?

Thanks!
Anson