What Is the Angle of Initial Velocity for a Projectile at Half Maximum Height?

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SUMMARY

The angle of the initial velocity vector for a projectile at half its maximum height can be determined using the relationship between the vertical and horizontal components of velocity. At this height, the speed is three-fourths of the initial speed. The equations of motion for projectiles, specifically the vertical motion equation \(v^2 = u^2 - 2gh\), are essential for solving this problem. By analyzing the components of velocity and applying the correct kinematic equations, one can derive the angle definitively.

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  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of vector decomposition
  • Basic grasp of gravitational acceleration (9.81 m/s²)
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  • Explore the concept of maximum height in projectile motion
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to clarify concepts related to velocity and angles in projectile dynamics.

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Homework Statement



At half its maximum height, the speed of a projectile is three-fourth its initial speed. What is the angle of the initial velocity vector with respect to the horizontal


Homework Equations



any projectile equation


The Attempt at a Solution



?
 
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Split the total speed into a vertical and horizontal component.
u_horiz = V cos(angle) u_vertical = V sin(angle)
The horizontal component doesn't change and the vertical component obeys v^2 = u^2 - 2 g h
 
okay so i tried splitting the horizontal and vertical components, but in the end and it doesn't accomplish anything because you get 2 unknown variables, also how would you account for the givens, such as, at half the max height you have 3/4 the initial velocity?

also i have tried using the equation V^2=V(o)^2 + 2 a x
from that i set V to (3/4 V(o))
g to (9.81) and
x to (x/2)
in the end that doesn't work out either??
 

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