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


by bentley4
Tags: physics by example, projectile
bentley4
bentley4 is offline
#1
Jan31-12, 10:25 AM
P: 66


In this solution the t in the y-coordinate equation is substituted using the x-coordinate 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?
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probl14.jpg  
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Delphi51
Delphi51 is offline
#2
Jan31-12, 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.
physicsisgrea
physicsisgrea is offline
#3
Jan31-12, 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.

bentley4
bentley4 is offline
#4
Jan31-12, 03:03 PM
P: 66

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


Quote Quote by Delphi51 View Post
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.
Ok, now I understand. Thnx!


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