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Projectile Motion Help?

  • Thread starter passmorea
  • Start date
  • #1
3
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 dont know where to plug this number into. I could answer this problem if it gave me the trajectory angle, but it is unknown?)
 

Answers and Replies

  • #2
21
0
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.
 
  • #3
3
0
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. :)
 
  • #4
3
0
now looking at it, .02 doesn't look right?
 

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