help! mechanics 2morrow.

[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]

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.

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.
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]

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

I think that there's a mistake here

maximum height it 9.8 not 2.8

[cooldude]

I think that there's a mistake here

maximum height it 9.8 not 2.8

edited my post

[nid404]

yup 8.89 was what i was gettin...but i wasn't too sure 

[zxcvbnm]

is this a past paper question?

[neno]

hey i dont get the part after u found the velocity 6 and ur finding the distance why to u subtract 8

[cooldude]

hey i dont get the part after u found the velocity 6 and ur finding the distance why to u subtract 8

if ur referring to mine then thats because the ball A has a headstart of 8 m, i.e. it is 8m above the ball B at the time of projection

[anne]

a straight footpath makes an angle alpha with the horizontal. An object P of weight 1250N rests on the footpath.The coefficient of friction between the object and the footpath is 0.1 The least magnitude of a force acting up the footpath, which will hold the object at rest on the footpath is 50N. By treating the object as a particle, show that the value of alpha satisfies:
10sinaplha - 10cosalpha =0.4
can somebody do this question!!?..i tried doin this..but got sinalpha - cosalpha =0.4
the 10 was missing

[anne]

^^^^ 10sinalpha - cosalpha =0.4

[anne]

anybody home?? OMG 

[cooldude]

a straight footpath makes an angle alpha with the horizontal. An object P of weight 1250N rests on the footpath.The coefficient of friction between the object and the footpath is 0.1 The least magnitude of a force acting up the footpath, which will hold the object at rest on the footpath is 50N. By treating the object as a particle, show that the value of alpha satisfies:
10sinaplha - 10cosalpha =0.4
can somebody do this question!!?..i tried doin this..but got sinalpha - cosalpha =0.4
the 10 was missing

let coefficient of friction=c, and alpha=x
therefore R=1250cosx, F=cR
therefore 1250sinx-F-50=0
1250sinx-125cosx=50
now divide both sides by 125-.
ull get 10sinx-cosx=0.4

[anne]

thanx alot!!