A quick projectile motion problem. Please help me Thanks

[Miss1nik2]

A marble is thrown horizontally with a speed of 23.4 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 61.8 ° with the horizontal. From what height above the ground was the marble thrown?

So far I have found Ay= -9.8, Ax= 0, Vox=23.4, Voy=0. And Vx=23.4 (im not sure if I need that).

Any help will be greatly appreciated! Thank you!

 

[AshesToFeonix]

remember velocity is a vector too and can be treated like distances. Draw a triangle to represent the x and y velocities. if you set it up right you should be able to use the Tan funtion to find the y velocity component. I'm sure you can figure it out from there

 

[baba89]

Hi there, it looks like you are on the right track with your initial velocity components. To solve this problem, you can use the equations of projectile motion. Since the marble is thrown horizontally, we can ignore the vertical acceleration and use the equation x = Vx * t, where x is the horizontal distance traveled, Vx is the initial horizontal velocity, and t is the time.

We know that the marble travels a horizontal distance of x = 0 (since it starts and ends at the same horizontal position) and a vertical distance of y = h (where h is the height above the ground that we are trying to find). We also know that the time it takes for the marble to hit the ground is the same for both the horizontal and vertical motions, so we can set the two equations equal to each other:

Vx * t = x = 0
Voy * t - (1/2) * g * t^2 = y = h

Using the values you have provided, we can plug them into the equations:

23.4 * t = 0
0 - (1/2) * (-9.8) * t^2 = h

Solving for t in the first equation, we get t = 0. Solving for h in the second equation, we get h = 0.

This means that the marble was thrown from a height of 0 meters above the ground, which makes sense since it starts and ends at the same horizontal position. I hope this helps! Let me know if you have any other questions.