Solving for initial velocity Projectile Motion

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a rock thrown from a 20-meter high building at a 53-degree angle. The horizontal range equals the building's height, leading to the equations of motion: Vx = Vox, Vy = Voy - gt, x = xo + Voxt, and y = yo + Voy - (1/2)gt^2. The user attempts to calculate the time of flight and the initial velocity but encounters difficulties with the negative displacement due to gravity. The correct approach involves substituting the time into the vertical motion equation to solve for the initial velocity accurately.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions
  • Basic grasp of gravitational acceleration (g = 9.81 m/s²)
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  • Review the derivation of projectile motion equations
  • Practice solving similar problems involving angles and heights
  • Learn about the impact of air resistance on projectile motion
  • Explore advanced topics like parametric equations in physics
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Students studying physics, educators teaching projectile motion, and anyone interested in applying kinematic equations to real-world scenarios.

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Homework Statement


A rock is thrown from the top of a twenty meter high building at the angle 53 degrees above the horizontal. IF the horizontal range of the throw is equal to the height of the building, what is the speed of the rock? How long is it in the air? What's the velocity of the rock before it hits the ground?



Homework Equations


Vx = Vox
Vy = Voy - gt
x = xo + Voxt
y = yo + Voy - (1/2)gt^2


The Attempt at a Solution


I use the second one and set x = to 20 and solve for the time...

20 = 0 + Vcos(53)t
t = 20/Vcos(53)

I substitute into the y equation

20 = vsin(53) - 1/2g(20/vcos(53)^2

Here is where I think I'm making the big mistake... help! :(
 
Physics news on Phys.org
The displacement y is along g. So it must be -ve
So rewrite the equation and solve for v.
 

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