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A projectile is fired upwards at a speed of 100 m/s at an angle of 37 with the horizontal.it explodes into two parts at the highest point in the mass ratio 1:3 and the lighter one comes to rest. The distance from the point of projection where the heavier mass lands is (g=10m/s^2)

Physics
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|dw:1358017831169:dw|
At highest point v=100cos37=80m/s If total mass of body is m, then masses of two pieces will be m/4 and 3m/4. The lighter one comes to a rest . By momentum conservation, velocity of 3m/4=4*80/3=320/3 m/s This new projectile will now cover a horizontal distance from its point of formation i.e. the highest point in the trajectory of previous projectile ,s=v*sqrt(2h/g) where v=320/3 as found before. and h is the point of maximum height found by applying kinematics to the projectile of mass m. The total distance from point of projection =s+ half of the range of previous projectile. The range can be calculated just like maximum height i.e. apply kinematics to it. I find it convinient to remember the formulas for range and max height h=(v sintheta)^2/2g Range,r=v^2 sin(2*theta)/g here theta=37 v=100
|dw:1358018149270:dw|

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Other answers:

R is the range ..(X com) is the centre of mass of the system..
@Diwakar i also used Same Kind of ur Approach But Didnt Get the Result..
and..@Taufique i did nt get ur Approach
there is no external force so centre of mass will be same at intial and final so ,i have applied this concept..
Centre of mass Will Be near to Heavier Body....
and using the Formula for Range i calculated Pos ( CM) = 960 m..ryt ?
The c.m. will finally be at R. It will follow a parabolic trajectory
The answer shuld be1120m
@Yahoo! ..wait one minute
@Yahoo! ..what is the answer in your text book for this question
1120m
sorry yahoo (xcom) without explosion =4m *R xcom after explosion =m*R/2 + k*3m ..then equate both then you will get right answer
@ yahoo! ..i am sorry for this mistake ...
@Yahoo! ..here k =1121.47 m answer
No Problem bro
4mR = m*R/2 + k*3m hw R/2 ????????
here explosion is self explosion i.e no external force is here so we apply this method
Any Reason Behind it ?..
because small part is in rest and it will drop vertically downward at R/2 distance from origin
we also solve this problem using momentum conservation ...
If u Dont Mind..Can u Show ?
i appl that method..but didnt get the result
ok..|dw:1358020235037:dw|
|dw:1358020365409:dw| |dw:1358020404165:dw| and find k..
|dw:1358020473006:dw|
|dw:1358020525435:dw|
any query, yahoo!
Got it..! THXXX
now i am giving you a question which is similar to it ..you are interested to solve this
An isolated particle of mass m is moving in horizontal plane (X-Y),along x-axis,at a certain height above the ground.it suddenly exlodes into two fragment of masses m/4 and 3m/4.An instant later the smaller fragment is at y=15 cm .what will be the position of larger fragment at this instant ???
@Yahoo! ..solve this ..it will give you more concept on this type question
Thxx...Brother.....i will try to Solve it
ok... try it.
35 ??
@Taufique wat is the answer
y=-5cm
hw....?
HINT: the body is moving on the x axis that is there is no motion along y axis this means the momentum will be zero along y-axis at any time
Huh..Silly....Careless
Didnt See it was on x-axis.......@Taufique if it was y axis..then the answer would be 35 ryt ?
now find momentum along y-axis for this situation.. after the body will explode ..the momentum will be zero along Y-axis ..am i right?
No..i Dont think
Since it is moving..along Y-axis...it will have an Initial P
no in the question ,it has been given that the body is moving along horizontal direction i.e X-axis
If it was Along Y-Along
*Axis
then the momentum will be zero along x -axis at any time..because there is no component of velocity along Xaxis
where is your confusion???
Cleared...
|dw:1358022098651:dw|
y=k (let ) for 3m/4 mass |dw:1358022439653:dw|
Got...It
ok..
Can u Take a luk at mmy Maths Question

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