Solving Projectile Motion: Horizontal & Vertical Distances

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SUMMARY

The discussion focuses on solving projectile motion problems, specifically calculating the horizontal distance and vertical height of a projectile fired at a 60-degree angle with an initial speed of 300 m/s. The horizontal component of the initial velocity is determined to be 150 m/s, while the vertical component is 259.8 m/s. Using the constant acceleration formula, the time of flight is calculated to be 53 seconds, allowing for further calculations of displacement in the horizontal direction.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations for constant acceleration
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to apply kinematic equations in two dimensions
  • Explore the effects of air resistance on projectile motion
  • Practice solving real-world projectile motion problems using different angles and speeds
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This discussion is beneficial for physics students, educators, and anyone interested in mastering projectile motion calculations in classical mechanics.

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Homework Statement



A projectile is fired at an angle of 60 degrees above the horizontal with an initial speed of 300 m/s. Calculate A) the horizontal distance traveled and B) the vertical height attained in the first 6 s.



Homework Equations





The Attempt at a Solution


I did 300cos60=150 so Vi(x)=150 and 300sin60=259.8 so Vi(y)=259.8 but after that it seems like i don't have enough information to continue.
 
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you need to use constant acceleration formulae

we know:
Ux = 150ms-1
Uy = 259.8ms-1
a = -9.81ms-2

First you can calculate how long it's in the air for, we know that its trajectory is symmetrical about the peak, and at the peak it's y velocity is 0ms-1.

[tex]v = u + at[/tex]


[tex]t = \frac{v-u}{a}[/tex]

= [tex]\frac{0 - 259.8}{-9.81} = 26.48s[/tex]

For it to hit the ground again, it must have been in the air twice that time = 53s.

now you know for the x plane, a, t and u
find s (displacement)

[tex]s = ut + \frac{1}{2}at^2[/tex]

hint: half of that equation dissapears when you sub in the values.

You can do the rest yourself!
 
Last edited:

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