Finding distance given only initial velocity

  • Thread starter mistabry
  • Start date
  • #1
12
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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;!
 

Answers and Replies

  • #2
60
0
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?
 
  • #3
12
0
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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