How Do You Calculate Initial Velocity and Launch Angle in Projectile Motion?

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Homework Help Overview

The discussion revolves around calculating the initial velocity and launch angle of a projectile, specifically a ball launched from ground level that lands after 2.5 seconds at a distance of 30 meters. The problem is situated within the context of projectile motion and kinematics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to begin solving the problem. Some participants suggest separating the horizontal and vertical components of motion to find the initial velocity and angle. Others question the validity of their calculations, particularly regarding the initial velocity derived from the kinematic equations.

Discussion Status

Participants are actively exploring different approaches to the problem, with some providing hints about considering horizontal and vertical components separately. There is a recognition of potential errors in calculations, and guidance has been offered regarding the horizontal motion and its implications for the vertical component.

Contextual Notes

The discussion is constrained by homework guidelines that limit the type of assistance that can be provided, emphasizing hints rather than complete solutions. Participants are also operating under the assumption that air resistance can be ignored.

creativeone
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1. A ball launched from ground level lands 2.5 s later on a level field 30 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)



2. Known Kinematic Equations



3. I don't know how to start!
 
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As per the homework section guidelines I can only give hints but have you considered the horizontal and vertical components separately as once you work them out you can work out the resultant and then use a bit of trigonometry to find the angle.
 
Yes, however, when I solved for initial velocity using x=v0t+1/2at^2, I got -0.25s. That doesn't sound right.
 
creativeone said:
Yes, however, when I solved for initial velocity using x=v0t+1/2at^2, I got -0.25s. That doesn't sound right.

Thats not the way I worked it out,

Start with the horizontal component as it doesn't change because we are ignoring the effects of air resistance.

So it travels 30m horizontally in 2.5sec, therefore

$$V = \frac{distance}{time} $$
$$V = \frac{30}{2.5} $$
$$V = 12ms^{-1} $$

Then consider the vertical component,

The initial velocity is unknown.
The acceleration will be -g
The time taken to reach the apex of the parabola will be half the total time.
The final velocity at the apex will be 0.

Can you work out the initial vertical component from that?
 

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