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Projectile motion -- launched 60° above the horizontal

  • #1

Homework Statement


A projectile is launched at an initial speed of 30 m/s at an angle of 60° above the horizontal. Calculate the magnitude and direction of its velocity (a) 2s and (b) 5s after launch

Homework Equations


x=v*t in the x dimension
x= x+vt+(1/2)at^2


The Attempt at a Solution


I tried going through with the x dimension to solve for acceleration but the acceleration in the y is 9.8.
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


A projectile is launched at an initial speed of 30 m/s at an angle of 60° above the horizontal. Calculate the magnitude and direction of its velocity (a) 2s and (b) 5s after launch

Homework Equations


x=v*t in the x dimension
x= x+vt+(1/2)at^2


The Attempt at a Solution


I tried going through with the x dimension to solve for acceleration but the acceleration in the y is 9.8.
You need to be clear about which coordinate (x or y) represents the horizontal direction and which represents the vertical direction.

If x represents the horizontal direction, why do you think the projectile is accelerating in that direction?
 
  • #3
311
23
Given that x is horizontal and y is vertical, the starting point is to break the launch velocity into its horizontal vector (x) ( which is deemed to be constant velocity )
and its vertical vector (y) which comes under the influence of gravitational deceleration.
Ive attached a sheet, which won't solve your problem but might give you a start.
 

Attachments

  • #4
248
26
The velocity change will only take place in the y direction. Calculate the x component of velocity and the y component after 2 seconds and find their resultant.
 
  • #5
You need to be clear about which coordinate (x or y) represents the horizontal direction and which represents the vertical direction.

If x represents the horizontal direction, why do you think the projectile is accelerating in that direction?
Yep thank you I was going in both.
 
  • #6
Given that x is horizontal and y is vertical, the starting point is to break the launch velocity into its horizontal vector (x) ( which is deemed to be constant velocity )
and its vertical vector (y) which comes under the influence of gravitational deceleration.
Ive attached a sheet, which won't solve your problem but might give you a start.
Thank you very much that helped Immensely!
 
  • #7
The velocity change will only take place in the y direction. Calculate the x component of velocity and the y component after 2 seconds and find their resultant.
Than you very much this made it so much easier!
 

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