Finding distance given only initial velocity

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SUMMARY

The discussion focuses on calculating the maximum distance a clown can travel when shot from a cannon at a constant speed of 11.1 m/s. The key equations used include Vx = Vo*cos(theta) and Vy = Vo*sin(theta), with the time of flight Δt derived from the vertical motion equation. The maximum distance occurs at a launch angle of 45 degrees, confirmed through differentiation of the horizontal distance equation Δx. The participant successfully resolves their confusion and confirms the solution.

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Homework Statement


In a circus act, clowns are shot from a cannon at a constant speed vo = 11.1 m/s. The angle of launch may be varied. What is the maximum distance D that a clown may travel, if the landing pad is at the same height as the mouth of the cannon?

Knowns:
Vo=11.1 m/s
x = 0
y = 0
V2=0
y2=0

Homework Equations


Vx= Vo*cos(theta)
Vy= Vo*sin(theta)
Δy= Vosin(theta)Δt + .5aΔt^2

The Attempt at a Solution


So I first solved for Δt,
Δy = 0 = (11.1)sin(theta) + (.5)(-9.8)Δt
Δt = (2(11.1)sin(theta))/9.8

Next I attempted to solve for Δx,
Δx = (11.1)cos(theta) * Δt

Substituting for Δt,
Δx = (11.1)cos(theta) * (2(11.1)sin(theta))/9.8

And I got lost D;!
 
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You almost got it!
For the equation of the horizontal distance you got, you need to find when is it maximal.
Getting maximal and minimal points is through derivation. If you do derive, what is the angle that gives you the maximal distance?
 
Wanna know something funny? Common sense tells me that angle 45 shoots the farthest... I can't believe I didn't see it before. d/dx (Δx) = 0 >.< I didn't see that before. Thank you! I solved my problem :]
 

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