Projectile Motion Distance Calculation

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SUMMARY

The discussion focuses on calculating projectile motion distance with constant velocity. Participants emphasize defining initial positions, x0 and y0, as zero to facilitate solving for the projectile's horizontal (x) and vertical (y) positions at a specified time, specifically at 2.9 seconds. The approach is straightforward and relies on basic kinematic equations.

PREREQUISITES
  • Understanding of basic kinematics
  • Familiarity with projectile motion concepts
  • Ability to solve equations involving time and distance
  • Knowledge of initial conditions in motion analysis
NEXT STEPS
  • Study the kinematic equations for projectile motion
  • Learn how to apply initial conditions in motion problems
  • Explore the effects of varying velocity on projectile trajectories
  • Investigate the role of gravity in vertical motion calculations
USEFUL FOR

Students in physics, educators teaching motion concepts, and anyone interested in understanding the principles of projectile motion calculations.

chelsea01
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Homework Statement
A projectile is launched at ground level with an initial speed of 46.0 m/s at an angle of 25.0° above the horizontal. It strikes a target above the ground 2.90 seconds later. What are the x and y distances from where the projectile was launched to where it lands?
Relevant Equations
Horizontal Motion(ax = 0)
x = x0 + Vx t
Vx = V0x = Vx = velocity is a constant.

Vertical Motion(assuming positive is up ay = −g = −9.80m/s squared)
y = y0 +1/2 (v0y + vy)t
vy = v0y − gt
(3.39)
y = y0 + v0yt −1/2gt2
This based on velocity being constant. Define x0 and y0 to be zero and solve for the desired quantities.
 
Physics news on Phys.org
Welcome to the PF. :smile:

Looks good so far. You should be able to solve for x and y at time = 2.9s now...
 

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