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Optimal angle for projectile motion?
Im doing a physics project for school on projectile motion and we have to work out the optimal angle. I have done my results and they are all fine. However we also have to calculate the maximum distance that is possible to be achieved without wind resistance. In my calculations the maximum distance occurs at 40 degrees, until I reach an initial velocity of 5.27m/s and then it changes to an optimal angle of 45 degrees. Also im not using any angle in between the two angles. Im using the formula range=ucos(theta) x time of flight, and i worked out to give me an overall equation when I work out the time of flight using the formula x=usin(theta)t - 0.5gt^2 which gives me the overall equation of what I have inserted as an attachment. So what I am asking is have gone wrong somewhere in my calculations or is it just normal to have different optimal angles for different velocities. One last thing is I know in the formula i have inserted is that the gxy should be read as 2gxy however i cancelled it out because of the 0.5 in the distance equation.
2 Answers
- 7 years ago
Hello! I did the same a few days ago and easily prooved, that optimal angle is 45 degrees, no matter what the distance is. First, express range in terms of angle (well, you did it, but I don`t know, if you did it correct), then differenciate it and find angle when dR/d angle = 0.
- Anonymous7 years ago
I don't know if the equation is right, but yes, you should expect that different velocities will chance the angle, ONLY IF there is a difference in y.