Find Shot Velocity Components | Projectile Motion Homework Help

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a shot putter who releases a shot at a specific velocity and angle. Participants are tasked with determining the x and y components of the shot's velocity at the end of its trajectory, given the initial conditions and time of flight.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of initial x and y components of velocity, questioning the use of acceleration and time in determining final velocities.
  • Some participants express confusion regarding the signs of the y-component of velocity and the implications of acceleration on both x and y components.
  • There are attempts to clarify the relationship between horizontal and vertical motion, particularly regarding the effects of acceleration on each component.
  • Questions arise about the correct interpretation of the equations used for calculating final velocities and horizontal displacement.

Discussion Status

The discussion is ongoing, with participants providing insights and corrections to each other's reasoning. Some guidance has been offered regarding the use of initial velocities and the nature of horizontal versus vertical motion. However, there is no clear consensus on the final calculations or methods to be used.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. There is also a focus on ensuring that the assumptions made about acceleration and velocity components are clearly understood.

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Homework Statement



A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 degree above the horizontal. The shot hits the ground 2.08s later. You can ignore air resistance.What is the x-component of the shot's velocity at the end of its trajectory?
What is the y-component of the shot's velocity at the end of its trajectory?

Homework Equations



x = v initial cos theta time

The Attempt at a Solution



I managed to get the x and y component of acceleration.

but for this I would suppose its going to be 12cos51 2.08 = 15.71

I am lost...as this is wrong.
 
Last edited:
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The 2.08 should be s, not m/s right?

What is the initial x and y component?

What is the acceleration of the shot put?
 
x-component of the shot's velocity at the beginning of its trajectory

7.55m/s

y-component of the shot's velocity at the beginning of its trajectory
9.32

y-component of the shot's acceleration while in flight
-9.80
x-component of the shot's acceleration while in flight
0

from what I got thusfar.

now I need to find the x and y component of the shot's velocity at the end of its trajectory
 
Last edited:
Heat said:
x-component of the shot's velocity at the beginning of its trajectory

7.55m/s

y-component of the shot's velocity at the beginning of its trajectory
-9.32

y-component of the shot's acceleration while in flight
-9.80
x-component of the shot's acceleration while in flight
0

from what I got thusfar.

now I need to find the x and y component of the shot's velocity at the end of its trajectory

cool. you have everything you need... you have initial velocity... acceleration... time... you should be able to get final velocity.

EDIT: why do you have -9.32 instead of 9.32?
 
so the equation would be vx = v initial +gt

vx = 11.99 + (9.8)(2.08)

= 32.37

but now how do I get x and y, (I don't have degree for the final velocity)
 
Heat said:
so the equation would be vx = v initial +gt

vx = 11.99 + (9.8)(2.08)

= 32.37

No... vertical acceleration affects vertical velocity... horizontal acceleration affects horizontal velocity...
 
sorry , but i don't see where you are going with this...
 
Heat said:
sorry , but i don't see where you are going with this...

The x-component of acceleration is 0... so does the x-component of velocity change?
 
no...same for y also...
 
  • #10
Heat said:
no...same for y also...

Ah... but there is a y-component to acceleration... calculate the final y-velocity using initial y velocity and the y-acceleration.
 
  • #11
vx = v0x + ax*t

vy = v0y + ay*t
 
  • #12
vx = v0x + ax*t

vy = v0y + ay*t

vx = 7.55 + (0)(2.08)

vy = 9.32 + (-9.8)(2.08)

vx = 7.55

vy = -11.064

Which is correct, thank you learning physics.
 
Last edited:
  • #13
also,

How far did she throw the shot horizontally?

this is what I did

I decided to use the following equation: x-x initial = t/2 (vinitial + vx)

I know the vf and vi for x and y.

so I decided to do a^2 + b^2 = C^2

so the velocity for initial would be 11.99
for final it would be 13.39

so now its

x-0 = 2.08 /2 (11.99+13.39)

x= 26.39

so that would be distance right? but I am wrong :(
 
  • #14
you're making it too complicated... horizontal displacement, you only need horizontal velocity... you know the horizontal velocity, you have the time... you know acceleration is 0...
 
  • #15
I tend to do that.

x-0 = 7.55(2.08) + .5...

final being 15.70 :)
 

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