## Mandy_Nakamoto Group Title what does equation of trajectory means?? what do i get from tat equation?? 0_o? 2 years ago 2 years ago

1. Taufique Group Title

trajectory equation means the locus of the motion of any body.from this equation you can get the information of the motion of that body...

2. Taufique Group Title

in projectile motion ,you can get position of body,maximum height travelled by that body,range of the motion of body and etc.from trajectory equation.

3. sgholap100 Group Title

you can use the formula: $y = x \tan (\theta) - ( (g * x^2) / (2 v^2 \cos^2(\theta) )$ where: y= the y coordinate of the end of the projectile x= the x coordinate of the end of the projectile theta= angle of inclination (from the horizontal) of the initial velocity g= gravitational constant, 9.81 v= initial velocity for this formula, you will place the initial point of the trajectory at the coordinates (0,0) hope this helps..^_^ OR JUST VISIT THIS SITE: http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

4. Taufique Group Title

see the picture ,i defined trajectory equation not for projectile motion it is general defination for motion, if this is projectile motion then equation of trajectory is obviously y=xtan(theta)-((g*x^2)/(2v^2cos(theta)).and for another motion it will be different _different. if you have any query then ask me.i will hint you .

5. Taufique Group Title

@sgholap100 you are right for only trajectory equation in projectile sense...

6. sgholap100 Group Title

yes it is different for all other paths so,a trajectory is the path that a moving object follows through space as a function of time. The object might be a projectile or a satellite

7. Mandy_Nakamoto Group Title

so this trajectory equation is constant??

8. Taufique Group Title

@sgholap100 yes it is true that the trajectory equation is the function of time but as you gave in the above the equation of the trajectory ,in this there is no time because we can eliminate also it in the function with respect to another variable i.e Y in X variable..;

9. Mandy_Nakamoto Group Title

There is this question.. A particle is projected from a point O on a horizontal plane with speed 35m/s at an angle of elevation α , where tan α =2. Find the maximum height above the plane of the particle and the equation of its trajectory. The answer for the trajectory equation is 2x-0.02x^2.. it substitues all the info we have and leave what we dont have.. so the final is my trajectory equation... is tat how i solve it??

10. Taufique Group Title

@mandy trajectory equation is the locus of path and the position of the body depends on time.

11. Taufique Group Title

@mandy, i gave you equation of trajectory for projectile motion ,put the value of v =35m/sec and theta=2 inthis equation and get your trajectory path i.e Y=2x-0.02x^2... and for maximum height y will be maximum because this is verticle distance which istravelled by body..find the maximum value of y on differentiating method and put dy/dx equal to zero and get points and find which point is maximum point and put this point in the given equation ,and get maximum value of y hence this will be maximum height .

12. sgholap100 Group Title

What answer did u got?

13. Mandy_Nakamoto Group Title

u said tat y is the maximum height.. why do i still have to differentiate it??? can't i just use it as it is??

14. sgholap100 Group Title

its beacuse u will get the height in a equation form and not in a number

15. Mandy_Nakamoto Group Title

why equation??

16. Mandy_Nakamoto Group Title

if i differentiate i'll still have x as an unknown..

17. sgholap100 Group Title

yes but u will get dy/dx=2-2*0.02x put dy/dx =0 therefore the equation will now become 0=2-0.04x therefore 0.04x=2 x=2/0.04 x=50 now subtitute these value in equation y=2x-0.02x^2

18. Mandy_Nakamoto Group Title

you guys r rite!!!! I got it!!! ^^ thank you x3!!

19. sgholap100 Group Title

welcome mandy we r happy to help : )