Two dimensional motion with constant acceleration problem,

[alexkolb]

Basketball player covers 2.80m horizontally in a jump.
His center of mass moves through the space as following:
his center of mass is at elevation 1.02m when he leaves the floor,
it reaches a maximum height of 1.85 above the floor, and its at 0.900m

when he touches down again.

determine his flight "hang" time, angle of takeoff, horizontal and

vertical velocity components at takeoff.

Please help, i don't know where too start, seems like too much is

unknown...

--

I started by making a drawing, and i thought that since max height and

range are given ill go from there, but the motion is not symmetrical

So i wrote out the X and Y motion equations:

Xf = Vix * t = Vi cosA * t , where A is the unknown angle V unknown

speed, t unknows time
Xy = Vyx *t = Vi sinA * t + 1/2gt^2

But i can't solve systems because there are 3 unknowns and 2 equations

So i went back to max height and range again, i tried to figure out

what the range was for symmetrical movement, but i couldn't since i

dont know the angle, and even if i did i wouldn't know what to do with

the rest of the motion trajectory, so that's it, I am stuck here, i don't

even know what to do because it seems that to solve it i need either

angle or initial speed..

Help please...

 

[alexkolb]

so? nobody has a clue?

 

[kyle8921]

I understand your frustration with this problem. It does seem like there are a lot of unknown variables and it can be overwhelming to figure out where to start. However, there are a few key pieces of information that we can use to solve this problem.

First, we know that the player covers 2.80m horizontally in a jump. This gives us the range, or horizontal distance, of the jump. We also know the maximum height and the elevation at takeoff and landing. This gives us the vertical distance of the jump.

Next, we can use the equations for constant acceleration to help us solve for the unknown variables. The equations are:

Xf = Xi + Vix * t + 1/2 * ax * t^2
Vf = Vi + ax * t
Vf^2 = Vi^2 + 2 * ax * (Xf - Xi)

In these equations, Xf is the final position, Xi is the initial position, Vix is the initial velocity in the x direction, t is the time, and ax is the acceleration in the x direction. We can also use the same equations for the y direction, using Viy and ay for the vertical velocity and acceleration, respectively.

Using the given information, we can set up a system of equations to solve for the unknown variables. We know that the final position in the x direction is 2.80m, the initial position is 0m, and the initial velocity in the x direction is unknown. We also know that the final position in the y direction is 1.85m and the initial position is 1.02m. The initial velocity in the y direction is also unknown.

We also know that the acceleration in the y direction is due to gravity, which we can assume to be -9.8 m/s^2. Since the motion is in two dimensions, we can also assume that the acceleration in the x direction is 0 m/s^2.

Using these equations, we can solve for the unknown variables. The flight "hang" time can be found by setting the final position in the y direction equal to the initial position in the y direction and solving for t. This will give us the time the player spends in the air.

The angle of takeoff can be found by using the equation for the x direction velocity and setting it equal to the given horizontal velocity of the jump, 2.80m/t.