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triggerhorse · Mar 22, 2009, 03:58 PM

I know that an object is thrown at a certain initial horizontal speed (m/s). I need to calculate the object's velocity (magnitude and direction) after a certain amount of time (seconds.) How would I do this? What formulas would I use? :confused:

Perito · Mar 22, 2009, 04:47 PM

The only forces acting on the object are 1) air friction and 2) gravity. Use formulas relating to those forces.

triggerhorse · Mar 22, 2009, 05:31 PM

Originally Posted by Perito

The only forces acting on the object are 1) air friction and 2) gravity. Use formulas relating to those forces.

Hi! In other questions I have been doing in Physics 11, air friction hasn't been discussed. I think you consider it Nil. I think I will only start dealing with air friction later.

Anyway, would you be able to show me the steps of the question to get the answer. I can't even seem to get a start on the question.

For the first step, would you be able to use the vfy=voy+ayt formula and figure out voy?

The actual question is: a calculator thrown at initial horz speed of 24m/s and calculate velocity(mag &direction) after 4 seconds.

Thanks

Perito · Mar 22, 2009, 07:46 PM

You break it down into two problems. One is the horizontal component and one is the vertical component. The vertical component is only gravity since there is no initial horizontal speed

$D_y$ is the displacement in the y direction
$D_y = V_oT + \frac12AT^2$
but $V_o = 0$
So $D_y=\frac12AT^2$
$D_y=\frac12GT^2$

In this case A = G, the acceleration due to gravity $(9.80665 \frac{m}{s^2})$

In the horizontal direction: $D_x$ is the displacement in the x direction

$D_x = V_oT + \frac12AT^2$
but $A = 0$
so $\frac12AT^2=0$
So $D_x=V_oT$

At any time, T, you can calculate the X and Y components of the vector.

You can calculate the resultant vector (see the figure) from the Pythagorean theorem:

$D = \sqrt{D_x^2+D_y^2}$

The resultant vector tells you where the object is at any time (T).

(I hope I did that right. It's been a while.)