How long is discus in the air and the horizontal distance traveled

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a discus thrown at an angle of 42.0° with an initial velocity of 28.0 m/s. Participants are exploring how to determine the time the discus remains in the air and the horizontal distance it travels.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss different equations for calculating the time of flight and horizontal distance, including the use of Δt=-2Vsinθ / g and dh=-V²*sin2θ / g. There are questions about when to apply these equations versus using dh=Vh*Δt or d=Vi*t+ 1/2*at².

Discussion Status

Some participants have provided insights into the equations and their applications, while others express uncertainty about the appropriate contexts for using different formulas. There is an acknowledgment that multiple methods can yield results, but clarity on their usage remains a point of discussion.

Contextual Notes

Participants reference specific equations from course materials and express confusion regarding their application in the context of projectile motion. There is an indication that the problem may involve assumptions about the projectile returning to the same launch level.

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Homework Statement



A discus is released at an angle of 42.0° with respect to the horizontal and a velocity of 28.0m/s.

a)How long does it stay in the air?

b)What horizontal distance does it travel?

Homework Equations


dv=1/2*at2
dh=Vh*Δt
Kinetic equation d=Vi*t+ 1/2*at2
dh=-V2*sin2θ /g
sinθ=opp/hyp
cosθ=adj/hyp
Δt=-2Vsinθ / g

The Attempt at a Solution



hyp=28.0m/s

cos(42°)*28.0m/s=Vh= 20.8m/s
sin(42°)*28.0m/s=Vv= 18.7m/s

a) I used Δt=-2Vsinθ / g and solved for Δt=3.82s

b) Used dh=-V2*sin2θ /g to solve for dh= 79.6m

For part b) I don't know when you I am supposed to use dh=-V2*sin2θ /g equation and when I should just use dh=Vh*Δt or the d=Vi*t+ 1/2*at2 equation. If anyone could give me any tricks or explanations of when and why we use that specific equation that would be greatly appreciated. Thank you so much in advance!
 
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hi dani123! :smile:
dani123 said:
a)How long does it stay in the air?

b)What horizontal distance does it travel?

For part b) I don't know when you I am supposed to use dh=-V2*sin2θ /g equation and when I should just use dh=Vh*Δt or the d=Vi*t+ 1/2*at2 equation.

(what's the dh=-V2*sin2θ /g equation? :confused:)

if you can use dh=Vh*Δt or d=Vi*t+ 1/2*at2 then you certainly should :smile:
 
dh=-V2*sin2θ /g equation was given in my course book for finding the horizontal range of a projectile, launched at an angle.
Where V is the instantaneous projectile velocity,
θ is the angle of projection, and
g is the acceleration due to gravity which is equal to -9.80 m/s2
This equation can be used to find the horizontal range of projectile that returns to the same level from which it was launched (dv=0), provided that angle of projection, θ, and the instantaneous projectile velocity, V, are known.
 
ahh! :smile:

in that case, the answer to your original question …
dani123 said:
For part b) I don't know when you I am supposed to use dh=-V2*sin2θ /g equation and when I should just use dh=Vh*Δt or the d=Vi*t+ 1/2*at2 equation.

… is that both methods work, so you should use the quick method if you can remember it

(which I'm pretty sure you won't be able to in the exam :wink:)
 
Thank you!
 

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