Need to find final Height, Equation not Working

Click For Summary
SUMMARY

The discussion centers on the application of kinematic equations to determine the final height of a projectile. The user initially applied the equation Sf = S0 + V0Δt incorrectly, leading to an erroneous result. The correct approach involves using Δy = V0yt + 1/2(a)(t²), which accounts for gravitational acceleration. The participants clarify that the first equation is suitable for horizontal motion, where gravity does not influence the x-component.

PREREQUISITES
  • Understanding of kinematic equations
  • Knowledge of projectile motion concepts
  • Familiarity with trigonometric functions (e.g., sine)
  • Basic principles of gravity's effect on motion
NEXT STEPS
  • Study the derivation and applications of kinematic equations in projectile motion
  • Learn how to resolve initial velocity into components using trigonometry
  • Explore the effects of gravity on vertical motion in-depth
  • Practice solving problems involving both horizontal and vertical motion
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion and its equations.

Lapse
Messages
49
Reaction score
4

Homework Statement


upload_2019-2-17_11-18-21.png


Homework Equations


The kinematic equations--namely, Sf = S0 + V0Δt

The Attempt at a Solution


[/B]
I am a bit confused as it seems this problem is very straight-forward.
My known variables:
X0 = 0m
Y0 = 1.7m
Δt = 3.92s
V0 = 29m/s
Θ = 60ο
Yf = ?

So, I just use the above equation for the y-component:

Yf = Y0 + V0y(Δt)

Yf = 1.7m + 29sin(60) * 3.92s

My final answer is in the hundreds, but the correct answer is 24.85m. Can someone steer me in the right direction, and also tell me why the equation I used doesn't work?

Thanks so much!
 

Attachments

  • upload_2019-2-17_11-18-21.png
    upload_2019-2-17_11-18-21.png
    10.2 KB · Views: 606
Physics news on Phys.org
Okay, I solved the problem by using Δy = V0yt + 1/2(a)(t2)

My only question, then, is why doesn't the other equation I used work? And when should I use that other equation?

Thanks.
 
  • Like
Likes   Reactions: Zack K
You should use the form d=(vi)(t)+1/2(g)(t)^2, because you have to take into account that gravity will affect the height reached over the time period.
 
Lapse said:
Okay, I solved the problem by using Δy = V0yt + 1/2(a)(t2)

My only question, then, is why doesn't the other equation I used work? And when should I use that other equation?

Thanks.
I think it's because your equation assumes that the object travels in a linear path without the affect of gravity.
 
  • Like
Likes   Reactions: Lapse
Zack K said:
You should use the form d=(vi)(t)+1/2(g)(t)^2, because you have to take into account that gravity will affect the height reached over the time period.

Understood. So would the other equation only be used for finding horizontal distance since gravity doesn't affect the x-component?​
 
Lapse said:
Understood. So would the other equation only be used for finding horizontal distance since gravity doesn't affect the x-component?​
Exactly :)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K