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Shot Out of a Cannon

Date: 10/12/98 at 11:21:04
From: J.D. Langridge
Subject: General Physics

In a circus act, a performer is being shot out of a cannon whose 
muzzle is at an angle of 30 degrees with respect to the horizontal.  
The circus performer is supposed to land in a net 20.0 meters away 
horizontally, and 1.5 meters away vertically above the point of 
ejection. With what initial speed, Vo, must the performer leave the gun 
in order to land safely in the net?

I thought of using the equation, v^2 = Vo^2 + 2ax, with x being equal 
to the displacement, v being equal to final velocity, Vo being equal 
to initial velocity, and a being equal to the acceleration. I'm stuck 
and don't even know where to start. Please help! Thanks.

Date: 10/12/98 at 17:54:16
From: Doctor Rick
Subject: Re: General Physics

Hi. Speaking for myself, I was never good at remembering all the 
different formulas, and the hardest part is remembering exactly when 
to use each one. So I always start at the beginning.

The human cannonball moves in two dimensions - horizontally and 
vertically. Horizontally, he or she moves at constant velocity (we 
ignore air resistance). Vertically, he or she starts moving up, but is 
accelerated downward by gravity.

       ^                            *
       |                   *                 *
       |             *                             *
       |       *                                      *
       |    *                                    1.5 m |
       | *  30deg              20 m                    |
    /  /                                               |
  /  /_________________________________________________|___

  x = v_0x t

  y = v_0y t - (g/2) t^2

You know the direction of the initial velocity. You want to find the
magnitude of the initial velocity. You will need to use the formulas:

  v_0x = v_0 cos 30deg
  v_0y = v_0 sin 30deg

From the equation for x, you can get an equation for t in terms of x. 
Plug this equation into the equation for y in terms of t. This will 
give you an equation for y in terms of x. At x = 20 m, you know that 
y = 1.5 m. Using these values, you will have an equation to solve for 

That's an outline of what you have to do. I hope you can figure out the 
details. If not, let me know just what you were able to do, and what 
you think you should do next, and I can give more help.

- Doctor Rick, The Math Forum   
Associated Topics:
High School Physics/Chemistry
High School Trigonometry

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