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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)
 one year ago
 one year ago
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)
 one year ago
 one year ago

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TaufiqueBest ResponseYou've already chosen the best response.1
dw:1358017831169:dw
 one year ago

DiwakarBest ResponseYou've already chosen the best response.0
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
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
dw:1358018149270:dw
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
R is the range ..(X com) is the centre of mass of the system..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@Diwakar i also used Same Kind of ur Approach But Didnt Get the Result..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
and..@Taufique i did nt get ur Approach
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
there is no external force so centre of mass will be same at intial and final so ,i have applied this concept..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Centre of mass Will Be near to Heavier Body....
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
and using the Formula for Range i calculated Pos ( CM) = 960 m..ryt ?
 one year ago

DiwakarBest ResponseYou've already chosen the best response.0
The c.m. will finally be at R. It will follow a parabolic trajectory
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
The answer shuld be1120m
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
@Yahoo! ..wait one minute
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
@Yahoo! ..what is the answer in your text book for this question
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
sorry yahoo (xcom) without explosion =4m *R xcom after explosion =m*R/2 + k*3m ..then equate both then you will get right answer
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
@ yahoo! ..i am sorry for this mistake ...
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
@Yahoo! ..here k =1121.47 m answer
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
4mR = m*R/2 + k*3m hw R/2 ????????
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
here explosion is self explosion i.e no external force is here so we apply this method
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Any Reason Behind it ?..
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
because small part is in rest and it will drop vertically downward at R/2 distance from origin
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
we also solve this problem using momentum conservation ...
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
If u Dont Mind..Can u Show ?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
i appl that method..but didnt get the result
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
ok..dw:1358020235037:dw
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
dw:1358020365409:dw dw:1358020404165:dw and find k..
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
dw:1358020473006:dw
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
dw:1358020525435:dw
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
now i am giving you a question which is similar to it ..you are interested to solve this
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
An isolated particle of mass m is moving in horizontal plane (XY),along xaxis,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 ???
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
@Yahoo! ..solve this ..it will give you more concept on this type question
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Thxx...Brother.....i will try to Solve it
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@Taufique wat is the answer
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
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 yaxis at any time
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Didnt See it was on xaxis.......@Taufique if it was y axis..then the answer would be 35 ryt ?
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
now find momentum along yaxis for this situation.. after the body will explode ..the momentum will be zero along Yaxis ..am i right?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Since it is moving..along Yaxis...it will have an Initial P
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
no in the question ,it has been given that the body is moving along horizontal direction i.e Xaxis
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
If it was Along YAlong
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
then the momentum will be zero along x axis at any time..because there is no component of velocity along Xaxis
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
where is your confusion???
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
dw:1358022098651:dw
 one year ago

TaufiqueBest ResponseYou've already chosen the best response.1
y=k (let ) for 3m/4 mass dw:1358022439653:dw
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Can u Take a luk at mmy Maths Question
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
http://openstudy.com/users/yahoo!#/updates/50eee432e4b0d4a537cdad0d
 one year ago
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