Shot Out of a CannonDate: 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. y ^ * | * * | * * | * * | * 1.5 m | | * 30deg 20 m | /\-----------------------------------------------|------>x / / | / /_________________________________________________|___ 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 v_0. 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 http://mathforum.org/dr.math/ |
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