A long jumper runs down the runway and it is measured that her vertical velocity is 2.7 m/s and her vertical velocity is 6.6 m/s. what is her angle of takeoff?

To find the angle of takeoff, we can use trigonometry. The vertical and horizontal velocities can be thought of as the vertical and horizontal components of the overall velocity vector.

In this case, we are given the vertical velocity (v⊥) as 6.6 m/s and the horizontal velocity (v∥) is not provided. However, we know that the total velocity (v) is the hypotenuse of a right triangle formed by the vertical and horizontal velocities.

To find the horizontal velocity, we can use the fact that the total velocity is the hypotenuse, and the vertical velocity (v⊥) and horizontal velocity (v∥) are the legs of the right triangle. We can use the Pythagorean theorem to find v∥:

v = √(v∥² + v⊥²)

Plugging in the values:

2.7 = √(v∥² + 6.6²)

Simplifying the equation:

2.7² = v∥² + 6.6²

Solving for v∥²:

v∥² = 2.7² - 6.6²

v∥² = 7.29 - 43.56

v∥² = -36.27

Since the equation has a negative value, it means that there is no real solution for the horizontal velocity. However, this is likely an error since the velocity components cannot be negative.

Therefore, the given values for vertical and horizontal velocities are not consistent, and it seems to be an error in the question.