# Thread: Motion with Uniform Acceleration

1. ## 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!!

2. 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:

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

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 $\displaystyle t \approx 0.3616\ s$ which means that the stone starts to fall 0.64 m above the window.

3. 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 $\displaystyle d$ = distance from where the stone was dropped to the top of the window

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

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

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

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

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

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

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

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

$\displaystyle 24 = 4gt + g$

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

$\displaystyle d = \frac{1}{2}g\left(\frac{24-g}{4g}\right)^2$

4. 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.