Just joined, not expecting answer to question just how to do it

  • #1
Alright guys, I just signed up to this forum cause I heard from a friend that you guys can help. I dont really want a solution to the problem, I just need help guiding me towards the solution. The problem is...

A gymnast works out on a trampoline. At the instant that she leaves the trampoline, a point on her waist is 2.3 m above the floor and at the center of the trampoline. At that instant, the point has an upward velocity of 7.8m/s and a horizontal velocity of 3.0 m/s. Write equations that describe the subsequent motion of that point and find its maximum height.

All ive got so far (which is probably wrong is): Xo = 0m VoX = 3m/s Yo = 2.3m Voy = 7.8 m/s as well as ax= 3m/s^2 and ay = -9.8 m/s^2

If that is right what would be the function that I plug these values into, and if not, where did I make my error. Thanks for anyone that can get me a little guidance here!
 

Answers and Replies

  • #2
Päällikkö
Homework Helper
519
11
There is no acceleration in the x-direction.

The basic equation:
[tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]
 
  • #3
Well doesnt she go 3m/s on x axis, or is there no kind of friction in this problem?
 
  • #4
147
0
well, the only error I can see is the ax=3m/s^2. You have an initial horizontal velocity, but there is no horizontal accelaration. You might want to make use of the formula
[tex]s_{x}(t)=s_{o} + v_{o}t + \frac{1}{2}a_{o}t^2 [/tex]
where s is the position, v is the velocity and a is the accelaration.
 
Last edited:
  • #5
147
0
damn.... not fast enough at the latex writing yet...
 
  • #6
Päällikkö
Homework Helper
519
11
Ready2GoXtr said:
Well doesnt she go 3m/s on x axis, or is there no kind of friction in this problem?
Yes, but there certainly are no forces accelerating her movement in the horizontal direction.
 
  • #7
dmoravec said:
well, the only error I can see is the ax=3m/s^2. You have an initial horizontal velocity, but there is no horizontal accelaration. You might want to make use of the formula
[tex]s_{x}(t)=s_{o} + v_{o}t + \frac{1}{2}a_{o}t^2 [/tex]
where s is the position, v is the velocity and a is the acceloration.

so x = 0m + (7.8m/s) + 1/2 * (-9.8m/s^2) ? Im sorta confused how I take the 7.8 m/s and include it with gravity
 
  • #8
147
0
well, if you include the time variable as well as your initial hight of 2.3 m
y(t) = 2.3 m + 7.8 m/s*t + 1/2 * -9.8 m/s^2 * t^2
and the output y(t) will be your height at time t
 
  • #9
So its basicaly (original height) + (Upward velocity) + effects of gravity over time?
or something like that
 
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  • #10
Im getting something like 5.4 m was her max height?
 
  • #11
K i think that equation must be the answer, thanks for the help, i really dont see why they told me horizontal motion if i dont even need to use it lol.
 

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