# Kinematics: A stone is projected vertically upwards, argh!

• Dec 22nd 2012, 05:35 PM
Furyan
Kinematics: A stone is projected vertically upwards, argh!
Hello,

I am having considerable trouble getting my head around questions involving the SUVAT equations. I can do them if I split them into up and down parts but I'm supposed to be able to do then in single steps and I'm finding that really difficult.

Here is an example:

A stone is projected vertically upwards from a point A with speed ums-1. After projection the stone moves freely under gravity until it returns to A. The time between the instant that the stone is projected and the instant that it returns to A is $\displaystyle 3\dfrac{4}{7}$ seconds. Modelling the stone as a particle. Show that u = $\displaystyle 17\dfrac{1}{2}$.

It seems to me that while the stone is moving upwards then it's acceleration, a, is -9.8 ms-2, but once it has reached it's maximum height and starts falling under gravity it's acceleration is 9.8 ms-2. How can I use an equation with 'a' in it if it's sign is changing? Also if is consider 'u' to be positive then 'v', once the particle is falling, must be negative. I don't understand how that works with the equations of motion.

Any help would be very much appreciated.

Thank you
• Dec 22nd 2012, 06:13 PM
MarkFL
Re: Kinematics: A stone is projected vertically upwards, argh!
The acceleration does not change sign, it remains constant at $\displaystyle -9.8\frac{\text{m}}{\text{s}^2}$, if we define "up" to be in the positive direction.

The velocity will change sign though, being initially positive and remains so until it begins falling after reaching its maximum height.

We will have:

$\displaystyle h(t)=-4.9t^2+ut$

Now, we are told:

$\displaystyle h\left(\frac{25}{7} \right)=0$

$\displaystyle -4.9\left(\frac{25}{7} \right)^2+u\left(\frac{25}{7} \right)=0$

$\displaystyle -4.9\left(\frac{25}{7} \right)+u=0$

$\displaystyle u=\frac{49}{10}\cdot\frac{25}{7}=\frac{35}{2}=17.5$