Need to find final Height, Equation not Working

AI Thread Summary
The discussion revolves around solving a kinematic problem to find the final height of a projectile. The initial approach using the equation Sf = S0 + V0Δt was incorrect because it did not account for the effects of gravity on the vertical motion. The correct equation, Δy = V0yt + 1/2(a)(t^2), incorporates gravitational acceleration, leading to the accurate final height of 24.85m. The conversation clarifies that the initial equation is suitable for horizontal motion, where gravity does not apply. Understanding when to use each equation is crucial for solving projectile motion problems effectively.
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: 570
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 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 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 :)
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top