A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years 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)
anonymous
 4 years 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)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358017831169:dw

anonymous
 4 years 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358018149270:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0R is the range ..(X com) is the centre of mass of the system..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Diwakar i also used Same Kind of ur Approach But Didnt Get the Result..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and..@Taufique i did nt get ur Approach

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is no external force so centre of mass will be same at intial and final so ,i have applied this concept..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Centre of mass Will Be near to Heavier Body....

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer shuld be1120m

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Yahoo! ..wait one minute

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Yahoo! ..what is the answer in your text book for this question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry yahoo (xcom) without explosion =4m *R xcom after explosion =m*R/2 + k*3m ..then equate both then you will get right answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ yahoo! ..i am sorry for this mistake ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Yahoo! ..here k =1121.47 m answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.04mR = m*R/2 + k*3m hw R/2 ????????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here explosion is self explosion i.e no external force is here so we apply this method

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Any Reason Behind it ?..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because small part is in rest and it will drop vertically downward at R/2 distance from origin

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we also solve this problem using momentum conservation ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If u Dont Mind..Can u Show ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i appl that method..but didnt get the result

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok..dw:1358020235037:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358020365409:dw dw:1358020404165:dw and find k..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358020473006:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358020525435:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now i am giving you a question which is similar to it ..you are interested to solve this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0An 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 ???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Yahoo! ..solve this ..it will give you more concept on this type question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thxx...Brother.....i will try to Solve it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Taufique wat is the answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0HINT: 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Huh..Silly....Careless

anonymous
 4 years 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 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now find momentum along yaxis for this situation.. after the body will explode ..the momentum will be zero along Yaxis ..am i right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since it is moving..along Yaxis...it will have an Initial P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no in the question ,it has been given that the body is moving along horizontal direction i.e Xaxis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If it was Along YAlong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then the momentum will be zero along x axis at any time..because there is no component of velocity along Xaxis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where is your confusion???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358022098651:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y=k (let ) for 3m/4 mass dw:1358022439653:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can u Take a luk at mmy Maths Question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/users/yahoo!#/updates/50eee432e4b0d4a537cdad0d
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.