Projectile Motion Help: Finding Maximum Height and Components

Click For Summary
SUMMARY

The discussion focuses on calculating the maximum height and the horizontal and vertical components of a projectile motion problem involving a ball fired from a cannon. Given a total distance of 235 meters and a flight time of 47 seconds, the horizontal component (vx) is determined to be 5 m/s. The vertical component (vy) is calculated using the formula vf = vi + at, where at maximum height the final velocity (vf) is 0 m/s, leading to a maximum height of 460.62 meters. The calculations utilize standard physics equations for projectile motion.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of kinematic equations for projectile motion
  • Learn about the effects of angle on projectile trajectories
  • Explore advanced projectile motion problems involving air resistance
  • Investigate the use of simulation tools for visualizing projectile motion
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in understanding the calculations involved in projectile trajectories.

passmorea
Messages
3
Reaction score
0
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?)
 
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
4K