# New to forum, lets see if you can help

• Mar 28th 2010, 07:57 AM
j55
New to forum, lets see if you can help
Hi, im in the middle of extending SUVAT equations, im learning from a book so have no tutor to ask about this, i hope someone here can help :)

if a ball is thrown from 1.2m above ground to a overall height of 2.45m then S=2.45 and So=1.2 correct?

Now when the ball comes back down to be caught in the hand what is S and So?

• Mar 28th 2010, 08:09 AM
Anonymous1
Please clarify what \$\displaystyle S \$ and \$\displaystyle S_0\$ denote.
• Mar 28th 2010, 08:31 AM
dag
Quote:

Originally Posted by j55
Hi, im in the middle of extending SUVAT equations, im learning from a book so have no tutor to ask about this, i hope someone here can help :)

if a ball is thrown from 1.2m above ground to a overall height of 2.45m then S=2.45 and So=1.2 correct?

Now when the ball comes back down to be caught in the hand what is S and So?

I will assume that

\$\displaystyle S_{0} \$ = initial velocity
while
\$\displaystyle S\$ = the height at a particular time.

Assuming that is correct and assuming that the ball is being caught at exactly the same height in which it was thrown from then \$\displaystyle S\$=1.2. The initial height will always be 1.2 and will not change since it is a constant and not effected by time.
• Mar 28th 2010, 11:07 AM
j55
re
i thought So was initial position not velocity? in suvat equations U represents initial velocity.
im quite confused because im reading different things in different places.
• Mar 28th 2010, 11:49 AM
e^(i*pi)
I learnt u was initial velocity.

The question is rather confusing, I can only assume that the ball is thrown from the hand which is 1.2m above ground so surely they're reversed.
• Mar 28th 2010, 01:12 PM
j55
re
thats what i thought, so if the ball is falling from 1.25m (regardless of how far it is falling) the initial position is 1.25m which is So yes?