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An aerialist on a high platform holds onto a trapeze attached to a support by an 8.1 m cord. Just before he jumps, the cord makes an angle of 45° with the vertical. He releases the trapeze at the instant it is 0.7 m below its initial height. Calculate the angle that the trapeze cord makes with the vertical at this instant.

Answer :

Final answer:

To calculate the angle the trapeze cord makes with the vertical at a specific instant, we can use trigonometry. Given that the cord is 8.1m and forms an angle of 45 degrees with the vertical before the jump, we can find the vertical component of the cord's length using the sine function. When the trapeze is 0.7m below its initial height, the vertical component will change, and we can find the new angle by equating the vertical component to the difference in heights.

Explanation:

To calculate the angle that the trapeze cord makes with the vertical at the given instant, we can use trigonometry. Since the cord is 8.1m long and forms an angle of 45 degrees with the vertical just before the jump, we can determine the vertical component of the cord's length using the sine function. Sin(45)=vertical component/8.1m. Solving for the vertical component gives us 8.1m * sin(45) = 5.74m.

When the trapeze is 0.7m below its initial height, the vertical component of the cord will also change. Let's call the new angle formed between the cord and the vertical as 'x'. Since the vertical component of the cord should be equal to the difference in heights, we have 5.74m - 0.7m = (8.1m) * sin(x). Rearranging the equation and solving for 'x' gives x = sin^(-1)((5.74m - 0.7m)/8.1m) = sin^(-1)(0.658) = 41.0 degrees.

Therefore, the angle that the trapeze cord makes with the vertical at the instant when the trapeze is 0.7m below its initial height is 41.0 degrees.

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