Two acrobats, each of 50.0 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 36.0 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?

I have no idea how to even start this question......

It's a basic freefall equation:

First the total time they are in the air will remain the same because the only acceleration will be gravity. So you can calculate time with 0=(6.8sin(36))t-4.9t^2, as you would for any free fall.

The horizontal velocity will tell us how far apart the two swings. As such, the horizontal distance can be split into two parts, before they push off (v1) and after (v2).
v1 will be 6.8cos(36).
Since v2 is the velocity needed to negate one of the acrobats speed, (they're traveling at the same speed) v2=2(v1)
They push off at the vertex, so the time of each segment would be t/2.
So, the first distance traveled would be (v1(t/2)). The second segment would therefore be ((v2)(t/2)) or (2(v1)(t/2)). add them together and you get 3(v1)(t/2)=x
Sorry, no calculator at the moment. hope this helps

To solve this problem, we can break it down into several steps. Let's go through each step one by one:

Step 1: Find the maximum height reached by the acrobats
The initial velocity and angle of launch can be used to find the vertical component of the initial velocity (V0y). Use trigonometry to calculate this:

V0y = V0 * sinθ
= 6.80 m/s * sin(36.0°)

Step 2: Calculate the time taken to reach the maximum height
We know that the time taken to reach the highest point of the trajectory (t_max) is equal to half the total time of flight (t_total). Since the acrobats start and end at the same height, the total time of flight will be doubled when compared to reaching the maximum height.

t_total = 2 * t_max

Step 3: Calculate the maximum height
The maximum height can be determined using the following kinematic equation:

y_max = V0y^2 / (2 * g)

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Step 4: Find the horizontal displacement during the ascent
The horizontal displacement (x_asc) during the ascent can be calculated using the following equation:

x_asc = V0 * cosθ * t_max

Step 5: Find the time taken to reach the same height as the starting swing
Since the second acrobat starts at the same height as the first swing, the time taken to reach the same height (t_same_height) will be the same as the time taken during the ascent (t_total).

t_same_height = t_total

Step 6: Find the horizontal displacement during the descent
The horizontal displacement (x_des) during the descent is given by:

x_des = V_same_height * cosθ * t_same_height

where V_same_height is the final velocity at the same height.

Step 7: Find the distance between the swings
The total distance (d_total) between the two swings is the sum of the horizontal displacements during the ascent and descent.

d_total = x_asc + x_des

By following these steps and plugging in the given values, you should be able to calculate the distance (in meters) between the two swings.