Projectile Motion: How to Calculate Velocity and Position Over Time

[FCPancakeIII]

This is just for fun but

I have a projectile launched at 35 degrees with a force of 25 N.

I know cos35*25N gives me horizontal force of 20.5 N
and sin35*25N gives me vertical force of 14.3 N

I've got my kinematics but it's been 3 quarters since I've done this so my question is:

How do I figure out the velocity(vertical I guess, since horizontal is constant) of the particle as a function of time. And I also forget how to turn that into a position versus time graph.

halp!

 

[Andrew Mason]

This is just for fun but

I have a projectile launched at 35 degrees with a force of 25 N.

I know cos35*25N gives me horizontal force of 20.5 N
and sin35*25N gives me vertical force of 14.3 N

I've got my kinematics but it's been 3 quarters since I've done this so my question is:

How do I figure out the velocity(vertical I guess, since horizontal is constant) of the particle as a function of time. And I also forget how to turn that into a position versus time graph.

halp!

You have to determine the acceleration. You have given us the force. What additional information do need in order to determine the acceleration?

AM

 

[FCPancakeIII]

I have the mass of the object, let's say 1kg. So I know the vertical and horizontal accelerations... but I don't know how to get from that to velocity.

 

[DocZaius]

Assuming no friction:

You need to know for what length of time or length of space the force was applied to the projectile. From there you can know its change in momentum or change in kinetic energy. Knowing the projectile's mass you will know its velocity.

From there, you can use kinematics to track x,y velocities and positions.

 

[Zula110100100]

I guess what your looking for is probably s=v_ot + 1/2at²
so the vertical displacement at a given time is

s = 14.3(t)+1/2(-9.8)(t²)

at one second it would be 9.4m
2 = 9m
2.91 = 0.

I got 2.91 by putting a 0 for s

0 = 14.3t-4.9t²
-14.3t = -4.9t²

divide by -4.9t

2.92 seconds = t (rounded)

Horizontally it would go 59.83m (rounded) (2.92*20.5), assuming it lands even with the launch