# Motion with Uniform Acceleration

• October 4th 2009, 02:27 AM
NathanBUK
Motion with Uniform Acceleration
Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.

Thankyouu!! :)
• October 4th 2009, 04:44 AM
earboth
Quote:

Originally Posted by NathanBUK
Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.

Thankyouu!! :)

You have to combine 2 equations:

$\left|\begin{array}{rcl}s+3&=&\frac12 \cdot 9.81 \cdot (t+0.5)^2 \\ s&=& \frac12 \cdot 9.81 \cdot t^2 \end{array} \right.$

Expand the first equation and subtract columnwise the second equation from the first one. You'll get an equation in t. Solve for t.

I've got $t \approx 0.3616\ s$ which means that the stone starts to fall 0.64 m above the window.
• October 4th 2009, 05:01 AM
skeeter
Quote:

Originally Posted by NathanBUK
Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.

Thankyouu!! :)

I'm assuming the stone was released from rest.

the $d$ = distance from where the stone was dropped to the top of the window

$t$ = elapsed time from when the stone was released until it gets to the top of the window

$d = \frac{1}{2}gt^2$

$d+3 = \frac{1}{2}g(t+0.5)^2
$

substitute $\frac{1}{2}gt^2$ for $d$ in the second equation ...

$\frac{1}{2}gt^2 + 3 = \frac{1}{2}g(t+ 0.5)^2$

$\frac{1}{2}gt^2 + 3 = \frac{1}{2}g(t^2 + t + 0.25)$

$\frac{1}{2}gt^2 + 3 = \frac{1}{2}gt^2 + \frac{1}{2}gt + \frac{1}{8}g$

$3 = \frac{1}{2}gt + \frac{1}{8}g$

$24 = 4gt + g$

$t = \frac{24-g}{4g}$

$d = \frac{1}{2}g\left(\frac{24-g}{4g}\right)^2$
• October 4th 2009, 05:08 AM
NathanBUK
Oh thankyouu soo much. One things for sure I deff havent picked this up well in class that working out loooks sooo confusing but Im sure ill get the correct answer if I follow youre working out.