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2d Kinematics Problem Final Velocity

  • Thread starter Succession
  • Start date
  • #1

Homework Statement



http://imgur.com/fGYf0

Stuck on #3

Homework Equations



V = Vi + a*t
Y = (1/2)gt^2 + (ViY * t) + Yi

The Attempt at a Solution


I have found the vertical and horizontal components and the initial velocity.
ViX = 6.2
ViY = 9.922
Vi = 11.7
Yf = 0
Yi = 7
Xi = 0
X = ?
Vf = ?
a (acceleration of gravity) = 9.8
(i = initial, f = final)

Answers to # 1 and # 2 are 9.922 and 5.0227 respectively.
 

Answers and Replies

  • #2
Bump: Any help here?
 
  • #3
cepheid
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Welcome to PF Succession!

Bump: Any help here?
Part 3 asks you to find the magnitude of the velocity at the end of travel, just before the object hits the ground. Well, what is the horizontal component of this velocity, at the end of the trip? Hint: do any forces act in the horizontal direction?

Okay, how about the vertical component of the velocity at the end of the trip? How would you find that? Hint: what height is it falling from? You know the "initial" velocity when it starts falling (i.e. at the point of max height), the acceleration, and the distance that it falls. Can you think of an equation that would allow you to solve for the final velocity (at the end of the fall) given those three things?
 
  • #4
Welcome to PF Succession!



Part 3 asks you to find the magnitude of the velocity at the end of travel, just before the object hits the ground. Well, what is the horizontal component of this velocity, at the end of the trip? Hint: do any forces act in the horizontal direction?

Okay, how about the vertical component of the velocity at the end of the trip? How would you find that? Hint: what height is it falling from? You know the "initial" velocity when it starts falling (i.e. at the point of max height), the acceleration, and the distance that it falls. Can you think of an equation that would allow you to solve for the final velocity (at the end of the fall) given those three things?
Well the horizontal component of the initial velocity is 6.2, as given. I found the time (t) to be 2.578866342. Then I found Vy using Vy= gt + Viy, Vy = -15.35081607. Then I plugged my answer into a^2 + b^2 = c^2, Vy^2 + Vx^2 = V^2. -15.35081607^2 + 6.2^2 = V^2.
ended with -16.55558981. answer is still counted as incorrect, not sure where I went wrong, as one of my classmates used this same process and walked me through it and he got the answer right.
 
  • #5
cepheid
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Well the horizontal component of the initial velocity is 6.2, as given. I found the time (t) to be 2.578866342.
I'm not sure what time this is supposed to be, but if it is the time required to reach the max height (starting from the moment of launch), then it is wrong, and it is not the time required to fall back down to the ground either. EDIT: I see, it looks like that might be the sum of those two times. But you didn't need time at all. Since you solved for the height in part 2 (and by the way you need to add the 7 m to that answer) you could have just used the equation vf2 = vi2 + 2ad. This is what I was strongly hinting at with my previous post.

Then I found Vy using Vy= gt + Viy, Vy = -15.35081607. Then I plugged my answer into a^2 + b^2 = c^2, Vy^2 + Vx^2 = V^2. -15.35081607^2 + 6.2^2 = V^2.
ended with -16.55558981. answer is still counted as incorrect, not sure where I went wrong, as one of my classmates used this same process and walked me through it and he got the answer right.
When you arrive at an answer, always always ask, "does this actually make any sense?" The magnitude of something can never be negative. What's more you definitely shouldn't have gotten a negative answer using Pythagoras, because the square of any number is always positive.
 

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