I with the formula for this marble falling from a platform

[DThes]

A marble is released at 22.5 meters per second from a 70 meter tall platform. When will the marble strike the ground if the equation for the platforms height is 2.3[t][/2]+ (-)6.5t+70?

 

[CWatters]

Welcome to the forum. Please don't delete the Homework template as it's there for a good reason.

In what direction is the marble released?

You have an equation for the height/position of the platform. Do you know how to get an equation for the velocity of an object from the equation for it's position?

 

[FactChecker]

Your question confuses me. Is the platform at a fixed height of 70 meters or is it at a changing height of 2.3[t][/2]+ (-)6.5t+70? Also, why would the platform height matter after the marble is released?

 

[DThes]

Welcome to the forum. Please don't delete the Homework template as it's there for a good reason.

In what direction is the marble released?

You have an equation for the height/position of the platform. Do you know how to get an equation for the velocity of an object from the equation for it's position?

Is it vx= dxf/t?

 

[DThes]

Your question confuses me. Is the platform at a fixed height of 70 meters or is it at a changing height of 2.3[t][/2]+ (-)6.5t+70? Also, why would the platform height matter after the marble is released?

I am not sure if the platform is at a fixed height. That's the whole the whole problem.

 

[CWatters]

I am not sure if the platform is at a fixed height.

No the position is varying according to that equation. Note that when you substitute t=0 you get the initial height of 70m. At other values of t you get a different height.

Is it vx= dxf/t?

Yes to go from position to velocity you differentiate the position equation. I suggest you do that and post an equation for the velocity of the platform.

 

[CWatters]

Your question confuses me. Is the platform at a fixed height of 70 meters or is it at a changing height of 2.3[t][/2]+ (-)6.5t+70? Also, why would the platform height matter after the marble is released?

It's not the height that matters it's the velocity. I assume the launch velocity (22.5m/s) is relative to the moving platform.

Correction: The height at launch will matter but it's known to be 70m.

 

[FactChecker]

It's not the height that matters it's the velocity. I assume the launch velocity (22.5m/s) is relative to the moving platform.

Correction: The height at launch will matter but it's known to be 70m.

Oh, of course. Thanks.

 

[jbriggs444]

Is it vx= dxf/t?

I assume that we are talking about an upward vertical release. So we are talking about vertical position -- aka height. We have the height of the platform as a function of time. We do not have anything that would aptly be called "x" or "f".

How about expressing the platform's vertical velocity as $v_p(t) = \frac{d\ h_p(t)}{dt}$