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

[Ready2GoXtr]

Alright guys, I just signed up to this forum cause I heard from a friend that you guys can help. I don't 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 I've 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!

 

[Päällikkö]

There is no acceleration in the x-direction.

The basic equation:
$$x = x_0 + v_0t + \frac{1}{2}at^2$$

 

[Ready2GoXtr]

Well doesn't she go 3m/s on x axis, or is there no kind of friction in this problem?

 

[dmoravec]

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
$$s_{x}(t)=s_{o} + v_{o}t + \frac{1}{2}a_{o}t^2$$
where s is the position, v is the velocity and a is the accelaration.

 

[dmoravec]

damn... not fast enough at the latex writing yet...

 

[Päällikkö]

Well doesn't 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.

 

[Ready2GoXtr]

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
$$s_{x}(t)=s_{o} + v_{o}t + \frac{1}{2}a_{o}t^2$$
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) ? I am sort of confused how I take the 7.8 m/s and include it with gravity

 

[dmoravec]

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

 

[Ready2GoXtr]

So its basicaly (original height) + (Upward velocity) + effects of gravity over time?
or something like that

 

[Ready2GoXtr]

Im getting something like 5.4 m was her max height?

 

[Ready2GoXtr]

K i think that equation must be the answer, thanks for the help, i really don't see why they told me horizontal motion if i don't even need to use it lol.