Two acrobats, each of 78.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 6.80 m/s and the angle of their initial velocity relative to the horizontal was 28.0 degrees upwards. At the top of their trajectory, their act calls for them to push against each in such a way that one of them becomes stationary in mid-air and then falls to the safety net while the other speeds away, to reach another swing at the same height at that of the swing they left. What should be the distance (in m) between the two swings?

Question

Two acrobats, each of 78.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 6.80 m/s and the angle of their initial velocity relative to the horizontal was 28.0 degrees upwards. At the top of their trajectory, their act calls for them to push against each in such a way that one of them becomes stationary in mid-air and then falls to the safety net while the other speeds away, to reach another swing at the same height at that of the swing they left. What should be the distance (in m) between the two swings?

Answer 1

To find the distance between the two swings, we need to calculate the horizontal distance traveled by one of the acrobats while the other is stationary.

Let's determine the initial horizontal velocity of the acrobats. We can do this by calculating the horizontal component of their initial velocity.

Horizontal component of velocity = Velocity x cosine(angle)

So, for one acrobat:
Horizontal velocity (v_x) = 6.80 m/s * cos(28.0°)

Now, since we want one acrobat to become stationary in mid-air, their horizontal velocity will become zero. We can use this information to find the time it takes for them to reach the top of the trajectory.

We can use the equation of motion:

v_x = v_0x + a_x * t

Since the initial horizontal velocity (v_0x) is given as 6.80 m/s and the horizontal acceleration (a_x) is zero (as they will become stationary), we can rearrange the equation to solve for time (t):

t = v_x / a_x
t = 6.80 m/s / 0 m/s²

The time it takes for the acrobat to reach the top of their trajectory will be zero since the horizontal acceleration is zero.

Now, let's calculate the vertical distance covered by the acrobat during this time.

We can use the equation of motion:

Δy = v_0y * t + (1/2) * a_y * t²

Since the initial vertical velocity (v_0y) is given as 6.80 m/s * sin(28.0°) and the vertical acceleration (a_y) is -9.8 m/s² (acceleration due to gravity), we can substitute these values into the equation:

Δy = (6.80 m/s * sin(28.0°)) * 0 + (1/2) * (-9.8 m/s²) * (0)²

The vertical distance covered by the acrobat during this time will be zero.

Therefore, the distance between the two swings should be zero meters since the acrobat becomes stationary and falls in mid-air while the other acrobat speeds away to the other swing at the same height.