Optimal Angle and Horizontal Range of a Bottle Rocket Projectile

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SUMMARY

The optimal angle for launching a bottle rocket to achieve maximum horizontal range is 45 degrees. This angle allows the projectile to utilize its initial velocity effectively, maximizing distance. Given a maximum height of 26.0 meters, the equations of motion can be applied to calculate the horizontal range. By substituting the time of flight derived from the vertical motion equation into the horizontal motion equation, the range can be determined accurately.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic knowledge of trigonometry
  • Ability to solve quadratic equations
NEXT STEPS
  • Calculate the time of flight for a projectile launched at 45 degrees using the height of 26.0 meters
  • Explore the derivation of the horizontal range formula for projectile motion
  • Learn how to apply kinematic equations to solve for projectile motion parameters
  • Investigate the effects of air resistance on projectile trajectories
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in the mechanics of launching projectiles effectively.

EmmaB03

Homework Statement


[/B]
A bottle rocket can shoot its projectile vertically to a height of 26.0m. At what angle should the bottle rocket be fired to reach its maximum horizontal range, and what is that range? (You can ignore air resistance).

Homework Equations



The Attempt at a Solution


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For the first part: I think the angle the rocket should be fired is 45 degrees, because it's the optimal angle for projectiles, where it travels the furthest when launched at this angle; am I right?

For the second part: I have no idea what to do to start this part of the problem. I know that I now have the angle (45) and the height (26.0). I don't know where to go from this point.
 
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I would begin with (and orienting our coordinate axes such that the initial position is at the origin ##\left(x_0,y_0\right)=(0,0)##):

##x(t)=v_0\cos(\theta)t\tag{1}##

##y(t)=-\dfrac{1}{2}gt^2+v_0\sin(\theta)t\tag{2}##

Solve (1) for ##t##, and substitute into (2)...what do you get?
 

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