3okie so dy/dx = 0 (maxima/mimnima condition)
a - 2bx =0 ...... x = a/2b ...............
[obuisly this is maxima only cuz minima is 0)
The equation of trajectory of a projectile is given by
y = ax - bx2
Find total distance travelled till the body reaches the highest point.
-
UP 0 DOWN 0 0 12
12 Answers
33oh!!if u want total distance and not only x then..........
ymax = a2/2b - a2/4b .......
= a2/4b .............
total distance = √x2+y2
62this is kind of standard..
you have to integrate ds
ds2 = dy2+dx2
dy=(a-2bx)dx
ds= √dy2+dx2= √(a2+1-2abx+4b2x2)dx
now this can be solved...
you limits of integration will have te be from x=0 to x=a/2b
62@iitimcoming this is nto the total displacement question..
you have to find the distance travelled :)
3hey ashish sorry that was a typo .. i first thought of velocity then i decided to do this!!!
62iitimcoming.. i dont think what you are doing is correct...
or I am not understanding what you are doing correctly.
3ohh... i got my mistake ... sorry didnt read the question properly[2]
