Projectile Motion- 2D Kinematics

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a mortar fired at a 60-degree angle, hitting a target 15 meters high and 30 meters away. The key equations utilized include the kinematic equations for horizontal and vertical motion: \(s = v_0 t + \frac{1}{2} a t^2\) and \(v = \frac{s}{t}\). The user correctly sets up equations for both the x and y components of motion, ultimately substituting time into the equations to solve for the launch velocity (VL). The final equation derived is \(15 = VL \tan(60) - 4.9 \frac{30^2}{(VL \cos(60))^2}\), which can be solved for VL.

PREREQUISITES
  • Understanding of 2D kinematics
  • Familiarity with projectile motion equations
  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Ability to manipulate algebraic equations
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  • Learn how to derive projectile motion equations in detail
  • Study the effects of different launch angles on projectile trajectories
  • Explore numerical methods for solving nonlinear equations
  • Practice additional problems involving projectile motion with varying initial conditions
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for problem-solving strategies in teaching these concepts.

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Homework Statement



A mortar elevated to 60 degrees scores a hit 15 m up a building 30 m away. With what speed was the shell fired?

I know I have to find the launch velocity, or VL. Here is a quick sketch i did:

nx50ts.jpg


Homework Equations




s= vot + 1/2at2

v = s/t

The Attempt at a Solution



I began by setting up a table of known factors:

2na43s4.jpg


I first set up a distance equation with the x values-

30 = (VL * cos60)t {the second part of the equation presumably cancels out, because acceleration is 0}

Then, I set up the equation using y values-

15 = (VL * sin60)t - 4.9t2

Next I tried solving for time using a different equation using x values:

v = s/t

t = 30/VL * cos60

This being said- do I just plug what I found for t into the x-distance equation? I feel as if I'm not doing the problem correctly because then I would just be squaring the velocity in the x direction and dividing by 30 which leads me back to the original distance. I apologize in advance if the answer is staring at me from right under my nose, I'm not the smartest student. Any and all help is greatly appreciated, thank you so much in advance!
 
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15 = (VL * sin60)t - 4.9t^2

Substitute the value of t = 30/(VL*cos60)

15 = VLtan60 - 4.9*30^2/(VLcos60)^2

Now solve for VL.
 

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