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 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
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)

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

Diwakar
 one year ago
Best ResponseYou've already chosen the best response.0At 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.04mR = m*R/2 + k*3m hw R/2 ????????

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

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

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

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

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

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

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

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

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

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

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

Taufique
 one year ago
Best ResponseYou've already chosen the best response.1An 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 ???

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

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

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

Taufique
 one year ago
Best ResponseYou've already chosen the best response.1HINT: 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

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

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

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

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

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

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

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

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

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

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

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