Kinematic and Projectile motion equations

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Homework Help Overview

The discussion revolves around a kinematics problem related to projectile motion, specifically focusing on finding the angle of projection where the maximum height is half of the range.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of SUVAT equations and share their progress in deriving the range formula. Questions are raised regarding the derivation of the maximum height formula and its relation to the range.

Discussion Status

Participants are actively engaging with the problem, sharing their findings and seeking clarification on specific equations. There is a collaborative effort to explore the relationships between range and maximum height, though no consensus has been reached yet.

Contextual Notes

Some participants express uncertainty about the derivation of specific equations related to maximum height and range, indicating a need for further clarification on these concepts.

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Homework Statement



Find the angle of the projection for which the maximum height is equal half of the range.


Homework Equations



Kinematic and Projectile motion equations
 
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Just use SUVAT equations.
 
I'm working on the same problem. Any help would be appreciated. I've gotten as far as finding R(range)= (Vo^2sin2Theta)/g but how do I find the formula for Max height?
 
LadyMario said:
I'm working on the same problem. Any help would be appreciated. I've gotten as far as finding R(range)= (Vo^2sin2Theta)/g but how do I find the formula for Max height?

[itex]R(range)= \frac{Vo^2 Sin2\Theta}{g}[/itex]

For maximum height
(v0Sinθ)2=2aR
 
How did you get that equation for max height?
 
LadyMario said:
How did you get that equation for max height?

(vy)2=(uy)2-2as
 
Thank you so much :)!
 

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