Projectile Motion Help: Finding Maximum Height and Components

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a ball fired from a cannon, with a total distance of 235 meters and a flight time of 47 seconds. Participants are tasked with finding the maximum height and the vertical and horizontal components of the motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between distance, time, and velocity, with one noting the horizontal component derived from the total distance and flight time. Others discuss the conditions at maximum height, questioning the application of kinematic equations and the role of initial vertical velocity.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning the validity of their results. Some guidance has been offered regarding the use of kinematic equations, but there is no explicit consensus on the correctness of the values obtained for the vertical component and maximum height.

Contextual Notes

Participants express uncertainty regarding the trajectory angle, which is not provided, and this affects their calculations. There is also a mention of potential errors in the computed vertical component and maximum height, indicating a need for further verification.

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A ball is fired out of a cannon and flies a total distance of 235 meters. The ball has a total flight time of 47s. What is the maximum height and the vertical and horizontal components?





Homework Statement



df= 235m
t= 47s

Homework Equations



d=vt
df=di+vit+(1/2)at^2
vf=vi+at
vf^2=vi^2+2a(df-di)

The Attempt at a Solution



d=vt
235=47v
v=5 (i don't know where to plug this number into. I could answer this problem if it gave me the trajectory angle, but it is unknown?)
 
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d=vt
235=47v
v=5

ok so u basically have the horizontal component here. if u put subscripts, it is dx = vx*ttot
therefor vx = 5m/s

ok next, you know at in projectile motion, at max height velocity = 0m/s right?
so vf = vi + at. t = 1/2 the total time because tup = tdown for projectile motion of the same height. remember this is all in the y component, so a would = 9.8m/s^2. plug and solve for vi ( which is also vy ) and u have ur y component.

next for maximum height you have found you vy component so what do u do? find another formula! what do we know again?? v at max height = 0m/s right? and you have vi sooooo which formula includes vf <--- v at max and vi AND DISTANCE IN THE Y DIRECTION? you already know. plug and chug and u shuld have it.
 
ok, so the horizontal component is 5. I didn't know where the 5 went.

and just checking my answers, did your vertical component equal .02

and also my max height was 460.62m.

thanks. :)
 
now looking at it, .02 doesn't look right?
 

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