Projectile Motion given a vertical distance and angle at that point

In summary, the conversation discusses solving a physics problem involving projectile motion and the use of the "Famous Five" equations for uniform accelerated motion. The key to solving the problem is recognizing that Vx and Vy are equal and using this information, along with known accelerations, to set up an expression for solving for more information.
  • #1
whoareyou
162
2

Homework Statement


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Homework Equations


Famous Five equations for uniform accelerated motion

The Attempt at a Solution


To be honest, I have absolutely no idea where to start. The only thing that I notice is that at the point where you're given the velocity vector (when the projectile is 10m above the ground) it makes an angle of 45 degrees, which means that both vx and vy are equal. I don't really know where to go from there.
 
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  • #2
whoareyou said:
To be honest, I have absolutely no idea where to start. The only thing that I notice is that at the point where you're given the velocity vector (when the projectile is 10m above the ground) it makes an angle of 45 degrees, which means that both vx and vy are equal. I don't really know where to go from there.

I'm glad you recognize Vx and Vy are equal. That is key to being able to solve the problem. :smile:

Since you know Vx and Vy are equal, think about what else you know.
What is the acceleration in the x-direction?
What about in the y-direction?

Knowing the starting position (just clearing the wall) can you use any of those "Famous Five" with Vx and Vy and your known accelerations to set up an expression that will let you solve for more information?
 
  • #3
There is no acceleration in the x direction, delta d x = vx delta t.
There is acceleration in the y direction, delta d = viy delta t + 1/2 g delta t squared

delta d y = delta d x x time + 1/2 g t squared --> i end up with -18 = 4.9t^2 which can't be done.
 
  • #4
NVM! got it YAY!
 
  • #5
whoareyou said:
NVM! got it YAY!

Alright then. Glad to hear it. :approve:
 

Related to Projectile Motion given a vertical distance and angle at that point

1. How do you calculate the initial velocity of a projectile given a vertical distance and angle at that point?

To calculate the initial velocity of a projectile, you can use the equation v = √(g*d/sin(2θ)), where g is the acceleration due to gravity (usually 9.8 m/s²), d is the vertical distance, and θ is the angle at that point. This equation is derived from the projectile motion formula y = y0 + (v0*sin(θ)*t) - (1/2*g*t²), where y0 is the initial height, v0 is the initial velocity, and t is the time.

2. How does the angle at which a projectile is launched affect its vertical distance?

The angle at which a projectile is launched affects its vertical distance because it determines the initial velocity and trajectory of the projectile. A higher angle will result in a longer vertical distance, while a lower angle will result in a shorter vertical distance.

3. Can a projectile reach the same vertical distance at different angles?

Yes, a projectile can reach the same vertical distance at different angles. This is because there are multiple combinations of initial velocity and angle that can result in the same vertical distance. However, the trajectory and time of flight may be different for each angle.

4. How does air resistance affect the vertical distance of a projectile?

Air resistance can decrease the vertical distance of a projectile. As the projectile moves through the air, it experiences air resistance, which is a force that opposes its motion. This force can decrease the velocity and therefore the vertical distance of the projectile.

5. Is the vertical distance of a projectile affected by the mass of the object?

No, the vertical distance of a projectile is not affected by the mass of the object. This is because the vertical distance is determined by the initial velocity and angle, which are independent of the mass of the object. However, the mass of the object can affect the time of flight and the horizontal distance of the projectile.

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