# Vertical velocity, Horizontal velocity, and Actual (resultant) Velocity of takeoff

1. Nov 6, 2011

### xgeek

1. The problem statement, all variables and given/known data
For my class we did a mini-lab that asked us to stand still and long jump. Kind of confused about how to solve because our teacher is horrible. I am supposed to calculate the vertical velocity at take off (which I thought was 0m/s but someone told me it's at the peak of the jump), the horizontal velocity of the takeoff, and the actual (resultant) velocity of the takeoff. I am supposed to calculate these values from the data I collected

displacement = 1.175 meters
time of flight = 0.49 seconds

I have some answers calculated trying to check them. Also supposed to get the angle of takeoff, but everyone in my class said that it is 45 degrees...

Thanks for taking your time to read this! Problem is for 10 extra points on our next test, well there will be a similar bonus question and I want to be able to understand how to do it correctly. Hopefully the values I collected will match some of the help I get. Thanks!

2. Relevant equations
vf=(vi)+(a)(t)
vf^2=(vi)^2+(2)(a)(d)
d=(vi)(t)+(1/2)(a)(t^2)

3. The attempt at a solution
vertical velocity at takeoff I got = 2.40 m/s
horizontal velocity at takeoff= 2.40 m/s
actual (resultant) velocity at takeoff= 3.39 m/s
angle of takeoff = 45 degrees

Last edited: Nov 6, 2011
2. Nov 6, 2011

### Staff: Mentor

Re: Vertical velocity, Horizontal velocity, and Actual (resultant) Velocity of takeof

You'll have to show how you calculated your results before we can comment (beyond saying that they don't appear to be correct...)

3. Nov 6, 2011

### xgeek

Re: Vertical velocity, Horizontal velocity, and Actual (resultant) Velocity of takeof

messed up. I got 2.40 m/s for the horizontal velocity, and also got 2.40 m/s for the vertical velocity. then I took A^2 + B^2 = C^2 to get resultant velocity. cant remember what i wrote on my paper, 3.39 m/s maybe? from there I did sin-1(2.40/resultant) and got 45 degrees

all from memory

Last edited: Nov 6, 2011
4. Nov 6, 2011

### Staff: Mentor

Re: Vertical velocity, Horizontal velocity, and Actual (resultant) Velocity of takeof

If the vertical and horizontal components of the velocity are essentially equal, then the angle should be very close to 45 degrees. $tan^{-1}\left(\frac{v_y}{v_x}\right)$.

5. Nov 6, 2011

### xgeek

Re: Vertical velocity, Horizontal velocity, and Actual (resultant) Velocity of takeof

Horizontal Velocity
1.175 = Vh (.49) + (1/2)(0.0)(.49)^2 <-horizontal is constant therefore a = 0
1.175 = Vh (.49)
2.397 m/s = horizontal velocity

vertical displacement = (0.0)(0.245) + (.5)(-9.81)(0.245)^2
dv = 0.294 m

Vertical Velocity
0.0^2 = Vi + (2)(-9.81)(.294)
2.40 is the vertical velocity

then Pythagorean theorem for the resultant 3.39 m/s

sin-1(2.40/3.39) = 45.069 degrees

6. Nov 6, 2011

### JHamm

Re: Vertical velocity, Horizontal velocity, and Actual (resultant) Velocity of takeof

The time it takes to fall from a height 'h' under the influence of gravity is:
$$T = \sqrt{\frac{2h}{g}}$$
Since the path you took as you flew was a parabola the time it took to reach your max height is equal to the time it took you to fall which is given by the formula above; from this you can calculate your max height and from this you can calculate the vertical part of your velocity.
From the period you can also calculate the horizontal part of your initial velocity since you know time and displacement.
The total velocity is the square root of the sum of the squares of the vertical and horizontal parts.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?