Maximum range of a projectile when launched at 45°(Physics by example, Rees, pr. 14)by bentley4 Tags: physics by example, projectile 

#1
Jan3112, 10:25 AM

P: 66

In this solution the t in the ycoordinate equation is substituted using the xcoordinate equation and ultimately leads to the answer. My questions: 1. Why don't I get the same answer when I substitute the v or v and t instead? 2. How am I supposed to know to substitute t in this example and not v? 



#2
Jan3112, 12:04 PM

HW Helper
P: 3,394

I don't see how you can get it by keeping the t. You don't know t, so how can you tell the maximum value of x that way? I get
x = gtē/(2*tanΘ) and can't tell what combination of t and Θ provide the maximum x. It would be good to see your calc. There is no need to eliminate the v; it is a constant. But you must eliminate the variable t. 



#3
Jan3112, 02:53 PM

P: 25

Well, the problem is quite easy to approach.
Range is given by u^2 sin(2a) / g, where a is the projection angle. Since 1<sin a<1, the max. value for a sine function = 1. This occurs when the angle is 90 degrees or .5pi radians. So, for a fixed u: 2a = 90 a = 45 degrees. 



#4
Jan3112, 03:03 PM

P: 66

Maximum range of a projectile when launched at 45°(Physics by example, Rees, pr. 14) 


Register to reply 
Related Discussions  
Physics projectile problem  finding vertical range using the horizontal range  Introductory Physics Homework  1  
Half of maximum possible projectile range  Introductory Physics Homework  2  
Horizontal Range and Maximum Height of a Projectile  Introductory Physics Homework  3  
projectile maximum range PLEASE HELP!  Introductory Physics Homework  5  
Finding time and range for a projectile launched from a cliff  Introductory Physics Homework  1 