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help! mechanics 2morrow.

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Light:
1)A ball is thrown vertically upwards with a speed of 14ms-1.  2 seconds later a second ball is dropped from the same point .find where the two balls meet.

nid404:
1)
Ohkay...a lil tricky here..

Ball 1

It's max height above the point can be found using v2=u2 + 2as

02=142+ (2X-10xs)

s=9.8m
time taken for covering this distance=1.4s  s=(u+v)/2 X t
2-1.4=0.6s
After 0.6 second from reaching it's max height, it falls back a distance of
s=1/2 X 10 X 0.62
 =1.8m
So it's displacement from the point is 9.8-1.8=8m

It's velocity at this point is given by v=u+at  v=0+10X0.6= 6m/s

At this point another ball 2 is dropped from 8m below the first ball (from the original point)

I'ma lil confused how to equate the two now....i need time to figure

Till then is someone can help it'd be great....Is this M1? Which paper?

cooldude:

--- Quote from: Light on May 18, 2010, 06:45:30 am ---1)A ball is thrown vertically upwards with a speed of 14ms-1.  2 seconds later a second ball is dropped from the same point .find where the two balls meet.

--- End quote ---

okay let the ball thrown with speed 14 m/s be ball A and the other ball B--->

let us start with time t=0 at the time when ball B is about to be projected i.e. 2 seconds after the projection of ball A-->

the time taken for A to reach its maximum height is-->
 
v=u+gt
0=14-10t
t=1.4 sec
the maximum height above the initial position wud be-->
s=14t-5t^2
s=9.8m
therefore there is still 0.6 sec left until ball B is projected downwards i.e. at t=0, now the displacement of A in these 0.6 sec wud be, s=0.5*10*(0.6^2), s=1.8 m, i.e. the distance from the initial projection point is 8m
the velocity of A at t=0, wud be-->
v=u+gt
v=0+(0.6*10)
v=6 m/s
therefore the displacement of A at t=0 wud be-->
s=6t+(0.5*10*t^2)-8
for B it wud be--> s=5t^2
now equate the distances-->
6t+5t^2-8=5t^2
t=4/3 sec
therefore after 4/3 sec the balls collide
s=5(t^2)
therefore s=5*16/9 m=8.89 m below the point of projection

yasser37:
I think I figured it out


so starting from where nid404    stopped

distance of one after 2 seconds is 8 above initial point
so first we need to find speed for the 1.8 seconds in the downward trip
V^2=2(10)(1.8)
V=6 -----> will use this a sinitial speed


S= 6t + 0.5 X 10 X t^2 - 8   -------------> where 8 is distance from initial position
and for ball 2
S=0.5 X 10 X t^2


solving simultaneously gives

t = 4/3

S = 5(4*3)^2

8.9 m below initial position

yasser37:

--- Quote from: cooldude on May 18, 2010, 08:04:55 am ---okay let the ball thrown with speed 14 m/s be ball A and the other ball B--->

let us start with time t=0 at the time when ball B is about to be projected i.e. 2 seconds after the projection of ball A-->

the time taken for A to reach its maximum height is-->
 
v=u+gt
0=14-10t
t=1.4 sec
the maximum height above the initial position wud be-->
s=14t-5t^2
s=2.8m
therefore there is still 0.6 sec left until ball B is projected downwards i.e. at t=0, now the displacement of A in these 0.6 sec wud be, s=0.5*10*(0.6^2), s=1.8 m, i.e. the distance from the initial projection point is 1m
the velocity of A at t=0, wud be-->
v=u+gt
v=0+(0.6*10)
v=6 m/s
therefore the displacement of A at t=0 wud be-->
s=6t+(0.5*10*t^2)-1
for B it wud be--> s=5t^2
now equate the distances-->
6t+5t^2-1=5t^2
t=1/6 sec
therefore after 1/6 sec the balls collide
s=5(t^2)
therefore s=5/36 m=0.139 m below the point of projection

--- End quote ---

I think that there's a mistake here

maximum height it 9.8 not 2.8

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