Projectile Motion maximum height

Click For Summary

Homework Help Overview

The discussion revolves around a projectile motion problem, specifically focusing on calculating the maximum height and initial velocity of a projectile. Participants are analyzing the equations of motion and the relationships between horizontal and vertical components of the projectile's trajectory.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to derive the initial velocity and maximum height using kinematic equations. There are discussions about the time of flight and the relationship between horizontal and vertical motion. Some participants question the correctness of their calculations and the assumptions made regarding the initial conditions.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and seeking verification from others. There is a lack of consensus on the correct answers, and multiple interpretations of the problem are being explored. Some participants have provided their calculated ranges for initial velocity and maximum height, while others express uncertainty about their methods.

Contextual Notes

Participants note discrepancies between their calculated values and the expected answers from an online assignment, indicating potential misunderstandings or miscalculations. There is also mention of additional height considerations based on measurement from the ground.

salman213
Messages
301
Reaction score
1
1. http://img525.imageshack.us/img525/6376/fic11p112xe1.png
http://img525.imageshack.us/img525/9550/vqc11p112el8.png
[/URL]



2. d = 1/2at^2 + v1t

v =d/t




3.
horizontal time it takes equals vertical time

Horizontal
t = d/v = 50 /(v1cos10)


Vertical
d = 1/2at^2 + v1t
-0.6 = 1/2(-9.81)(50/v1cos10)^2 + v1sin10(50 /(v1cos10)

i get v1 = 36.6 m/s

-1.6 = 1/2(-9.81)(50/v1cos10)^2 + v1sin10(50 /(v1cos10)
i get v1 = 34.8 m/s


So my range is 34.8 m/s to 36.6 m/s

for b part

dv = v1(t) - 1/2(9.81)(t^2)

maximum height occurs at half the time so i used 1/2(50/v1cos10)

dv = 34.8(1/2(50/36.64cos10) - 1/2(9.81)(1/2(50/34.8cos10))^2
i got 22.8 m

for 2nd velocity

dv = 36.64(1/2(50/36.64cos10) - 1/2(9.81)(1/2(50/36.64cos10))^2
i got 23.0 m


So my maximum height range is 22.8 m to 23.0 m

According to my online assignment this is wrong

please helppppppppppp

 
Last edited by a moderator:
Physics news on Phys.org
ALSO:

i just noticed for maximum height i guess I also add 2.1 m if they are measuring from the Ground so 22.8 + 2.1 = 24.9 m and 23.0 + 2.1 = 25.1 m but these values still are somehow wrong...
 
i know it may be a little long in calculating but can someone check to see if my process is right if no one wnats to check my calculations :)
 
Hi Salman, i have been doing some work on this problem, Would you please be so kind to include the correct answers so i can verify and post it?
 
if i had the correct answers i wouldn't have posted it :(

wat answers did u get? and how did u attempt this problem differently?

Thanks..
 
vipul did u get different answerS?
 
sorry salman, haven't got the answers yet but still working on them, i thought if u had the correct answers, maybe we can work backwards.
 
hmm ok if u get answers tell me because i can keep submitting the answers repeatedly on my assignemtnt o see if they are right but I need the right answer. I don't understand what I am doing WRONG!


also do u have a energy question? about a spring and stuff? Did u fnish that one :S?
 
Salman, for part (a), my answers; the range is between 34.99m/s to 36.82m/s. Could you please verify these?
 
  • #10
hmm its one question so try getting b part too about maximum height range,

like its ONE question

a)
b)

and u have to enter both answers.


then i can cehck if they are right,


if someone else can check my solution please do, i don't know where I am going wrong!
 
  • #11
Ok, my answers;

a) 34.99m/s to 36.82m/s
b) 3.99m to 4.18m
 
  • #12
lol its too late sry my assignments done
 
  • #13
Did u get the correct answers?
 

Similar threads

Replies
40
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
2K
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K