At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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...
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.
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
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 .
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
so this trajectory equation is constant??
@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..;
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??
@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 .
What answer did u got?
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??
its beacuse u will get the height in a equation form and not in a number
if i differentiate i'll still have x as an unknown..
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
you guys r rite!!!! I got it!!! ^^ thank you x3!!
welcome mandy we r happy to help : )