What is the Takeoff Angle for a Long Jumper?

  • Thread starter Thread starter bionut
  • Start date Start date
  • Tags Tags
    Angle Projectile
Click For Summary
The discussion focuses on calculating the takeoff angle for a long jumper who rises 0.5 meters with a forward velocity of 8 m/s. A participant initially misinterprets the velocity as the hypotenuse of a right triangle, but it should be considered the horizontal component. The vertical component of velocity is unknown, and the height of 0.5 meters represents maximum height, not velocity. To find the correct angle, one must use the appropriate trigonometric relationships involving both horizontal and vertical components. Clarifying these points is essential for accurately determining the takeoff angle.
bionut
Messages
54
Reaction score
0
"A long jumper rises 0.5 meters during the flight phase of his jump. His forward velocity is 8 ms. What is the angle of take off?"

Okay, I did a skectch to work out basic trig, with 8m/s as my H and 0.5m as my 0
using sinϑ=o/h, =0.5/8 = 3
but it should be 21.6 ... do you know where I went wrong??
 
Physics news on Phys.org
I see a couple of problems so far. The first is that I would interpret the 8 m/s as the horizontal component of the velocity. Therefore if you draw a right triangle consisting of the velocity vector and its two components, the 8 m/s is not the hypotenuse of the triangle, but rather the horizontal leg.

The second problem is that you don't know what the vertical component is. You have to be careful here -- the 0.5 m provided is a distance, not a velocity. It's a little ambiguous what the "flight phase" of the jump is, but if you assume 0.5 m to be the maximum height that he reached, then you have enough info to solve the problem.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
13K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
13K
  • · Replies 7 ·
Replies
7
Views
7K
Replies
2
Views
5K
Replies
3
Views
2K