# motion in one-d

• May 30th 2010, 02:11 PM
jiboom
motion in one-d
ball thrown up with v=10m/s from a building of height 100. find max height and time taken to reach this max.
i said $\displaystyle a=-g$

integrate to get $\displaystyle v=-gt+C$ but initially v=10 so $\displaystyle v=-gt+10$

at max height v=0 so t=10/g

now i integrate again to get $\displaystyle s=-(gt^2)/2+10+K$but initially s=100 so K=100

so at t=10/g, i get s= 5.1

this answer is correct but they use

$\displaystyle v_{0}^2-2g \Delta h=v$

can anyone show me how they got this equation?

For part b) they want the time taken to go from h=50 to h=0.

for this i used the formula above for s and solved for t wneh h-50 and 0 to get the correct answer.

in the solutions they use
$\displaystyle v_{0}^2-2g \Delta h=v$

and if v2 is velocity at H=0 and v1 is velocity at H=50 then

v2=v1-gt

is the above obvious?im thinking,wrongly, if i start motion at v1 then f=ma gives me a=g so v=gt+K but initially v=v1 thus v=gt+v1 ??

Can i get v2=v1-gt starting with f=ma?
• May 31st 2010, 04:35 AM
HallsofIvy
Quote:

Originally Posted by jiboom
ball thrown up with v=10m/s from a building of height 100. find max height and time taken to reach this max.
i said $\displaystyle a=-g$

integrate to get $\displaystyle v=-gt+C$ but initially v=10 so $\displaystyle v=-gt+10$

at max height v=0 so t=10/g

now i integrate again to get $\displaystyle s=-(gt^2)/2+10+K$but initially s=100 so K=100

This should be $\displaystyle s= -gt^2/2+ 10t+ K$.

Quote:

so at t=10/g, i get s= 5.1
What? a ball is thrown upward from the top of a 100 m building and you assert that its highest point it 5.1 m??

What value did you use for g? 9.8? $\displaystyle -9.8(1)/2+ 10(1)+ 100= 105.1$. If you are calculating "height above the building" then your "K" is 0.

Quote:

this answer is correct but they use

$\displaystyle v_{0}^2-2g \Delta h=v$

can anyone show me how they got this equation?
Did they really set that equal to v? From your $\displaystyle v= -gt+ v_0$, $\displaystyle t= (v_0- v)/g$. Put that into $\displaystyle s= \Delta h= (-g/2)t^2+ v_0t$ and you get $\displaystyle \Delta h= -(g/2)(v_0^2- 2v_0v+ v^2)/g^2+ v_0(v_0- v)/g= -v_0^2/2g+ v_0v/g- v^2/2g+ v_0^2/g- v_0v/g$$\displaystyle = v_0^2/2g- v^2/2g$. Multiply both sides by 2g to get $\displaystyle 2g\Delta h= v_0^2- v^2$ which is the same as $\displaystyle v^2= v_0^2- 2g\Delta h= 0$, not v.

You can also interpret that in terms of energy. At velocity v, the kinetic energy is $\displaystyle (1/2)mv^2$ and at height $\displaystyle \Delta h$ the potential energy is $\displaystyle mg \delta h$. Multiply both sides of $\displaystyle v^2= v_0^2- 2g\Delta h$ by m/2 and you have $\displaystyle (1/2)mv^2= (1/2)mv_0^2- mg\Delta h$ or $\displaystyle (1/2)mv^2+ mg\Delta h= (1/2)mv_0^2$- that the total energy, kinetic energy plus potential energy, at any point is equal to the kinetic energy at the start where the potential energy was 0.

Quote:

For part b) they want the time taken to go from h=50 to h=0.

for this i used the formula above for s and solved for t wneh h-50 and 0 to get the correct answer.

in the solutions they use
$\displaystyle v_{0}^2-2g \Delta h=v$

and if v2 is velocity at H=0 and v1 is velocity at H=50 then

v2=v1-gt

is the above obvious?im thinking,wrongly, if i start motion at v1 then f=ma gives me a=g so v=gt+K but initially v=v1 thus v=gt+v1 ??
a= -g doesn't it? so that v= -gt+ v1?

Quote:

Can i get v2=v1-gt starting with f=ma?
Yes, that's what I just said: v= -gt+ v1 is the same as v2= v1- gt.
• Jun 7th 2010, 01:18 AM
jiboom
Quote:

Originally Posted by HallsofIvy

What? a ball is thrown upward from the top of a 100 m building and you assert that its highest point it 5.1 m??

.

bit rushed typing it in, i do realise i need 100+5.01 :) Im bad at this stuff but not that bad yet

Quote:

Originally Posted by HallsofIvy
a= -g doesn't it? so that v= -gt+ v1?

just to verify, this is -g because we initially throw it up ? If i just dropped it then iwould choose the direction so a=g?
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