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

In summary, the discus is released at an angle of 42.0° with a velocity of 28.0m/s. The discus stays in the air for 3.82 seconds and travels a horizontal distance of 79.6 meters. There are multiple methods for solving for the horizontal distance (dh), including using dh=Vh*Δt and d=Vi*t+ 1/2*at2 equations, as well as the dh=-V2*sin2θ /g equation, which can be used if the angle of projection and instantaneous projectile velocity are known.
  • #1
dani123
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0

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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  • #2
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:
 
  • #3
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.
 
  • #4
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:)
 
  • #5
Thank you!
 

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

1. How is the time in the air of a discus calculated?

The time in the air of a discus is calculated using the equation t = 2Vsinθ/g, where V is the initial velocity of the discus, θ is the angle at which it is thrown, and g is the acceleration due to gravity.

2. Does the weight of the discus affect the time in the air?

Yes, the weight of the discus does affect the time in the air. A heavier discus will have a shorter time in the air compared to a lighter one, assuming they are thrown with the same initial velocity and angle.

3. What factors affect the horizontal distance traveled by a discus?

The horizontal distance traveled by a discus is affected by the initial velocity, angle of release, air resistance, and the height from which it is thrown.

4. How does air resistance affect the flight of a discus?

Air resistance, or drag, can have a significant impact on the flight of a discus. It creates a force that acts opposite to the direction of motion, slowing down the discus and reducing its horizontal distance.

5. Can the horizontal distance traveled by a discus be increased by changing the angle of release?

Yes, the horizontal distance traveled by a discus can be increased by changing the angle of release. Generally, a higher angle of release will result in a longer horizontal distance, as long as the initial velocity remains constant.

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