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Need help with Vertical Velocity I have an exam tomorrow >.<

  • Thread starter newgirl
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Need help with Vertical Velocity!! I have an exam tomorrow >.<

I went to a tutor yesterday to get it all worked out, but I forgot to write down D in my notes! How do I solve for D? I have the answer, but no equation to work it.

You jump horizontally with a speed of 2.7 m/s off a diving board that is 2.6 meters above the water. (Neglect air resistance and use g = 9.8 m/s2)

(a) How long are you in the air?
0.73 s

(b) How far away from the diving board (horizontally) do you hit the water?
1.971 m

(c) What is your horizontal speed when you hit the water?
2.7 m/s

(d) What is your vertical velocity when you hit the water? (Define up as the positive vertical direction).
-7.252 m/s

I am thinking D is solved with Dy=Voy T + (1/2)Ay T^2=Voy T- (1/2)gt^2. I just figured out that the y's and the x's are referring to vertical and horizontal, but I am slightly confused on the process.
 
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Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

I think you just consider that you're falling from h=2.6m so use PE = KE so that the PE=mgh
and the KE=1/2 m v^2 where v is the vertical velocity.
 
Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

I think you just consider that you're falling from h=2.6m so use PE = KE so that the PE=mgh
and the KE=1/2 m v^2 where v is the vertical velocity.
This is a physics non majors course, and I haven't seen PE=KE yet. Initial=final? From what I googled PE=KE is energy related. The equations he gave us to work with was the previous one and Vy=Voy + AyT=Voy-gt

http://www.tjhsst.edu/~jleaf/tec/html/10/potent.htm [Broken] that kinda seemed to explain what you said.
 
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Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

simply use v^2=u^2+2xaccelerationxheight/distance and solve for v which should get 7.14 m/s
 
Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

only use vertical velocity components negelct horizontal velocity
 
Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

simply use v^2=u^2+2xaccelerationxheight/distance and solve for v which should get 7.14 m/s
What does the U stand for? Sorry if I sound really lame, but I really know nothing about physics other than what the teacher lectures or what it says in the book.
 

Simon Bridge

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Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

It's not too unusual that you meet ballistics before you really get into conservation of energy. What you do is separate the velocity into components and apply the kinematic equations to each component separately. It is common to make the x-axis horizontal and the y-axis vertical.

If ##\theta## is the angle the velocity makes to the horizontal, then ##v_x=v\cos\theta## and ##v_y=v\sin\theta## and ##\vec{v}=v_x\hat{\imath}+v_y\hat{\jmath}##

Horizontally there is no force, so you just keep going. ##x=v_xt##
Vertically there is a force (gravity) so you need the kinematic equations - (the suvat equations) or you can sketch a v-t graph (this one's easy) and do some geometry (slope of a line and area of a triangle).

What does the U stand for?
google for "SUVAT equations"... it is a common way to write "initial velocity" in the kinematic equations.
 
Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

u stands for intial velocity, so in terms of your vertical velocity the diver starts of at a intial vertical velocity of 0m/s. but for the initial horizontal velocity 2.7 m/s (also marked u) is the same as the horizontal final velocity which is also 2.7m/s as you have stated in your answer.
 
10,686
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Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

This is a physics non majors course, and I haven't seen PE=KE yet. Initial=final? From what I googled PE=KE is energy related. The equations he gave us to work with was the previous one and Vy=Voy + AyT=Voy-gt

http://www.tjhsst.edu/~jleaf/tec/html/10/potent.htm [Broken] that kinda seemed to explain what you said.
The PE stands for potential energy so an object at h=2.6m has potential energy = m g h

as it falls it converts the PE to kinetic energy so that at h=0 ground level PE=0 and KE=(PE initially) and since KE=1/2m v^2 then you can solve for v.

The other way to solve it is to ignore the horizontal velocity for the moment and think okay I start with 0m/s falling and after falling 2.6 my vertical velocity would be vfinal = g t but I don't know what t is but I remember that distance fallen is 1/2 g t^2 ie 2.6=1/2 g t^2 and solve for t. having t I go back to vfinal = gt and use t to find vfinal.
 
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Re: Need help with Vertical Velocity!! I have an exam tomorrow >.<

Thank you so much everyone! I think I got it now. I'll bookmark all the helpful links!
 

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