A spring cannon is located at the edge of a table that is 2.00 m above the floor

Question

A spring cannon is located at the edge of a table that is 2.00 m above the floor. A steel ball is launched from the cannon with speed vi at 36.0° above the horizontal.
(a) Find the horizontal position of the ball as a function of vi at the instant it lands on the floor. We write this function as x(vi).
(b) Evaluate x for vi = 0.200 m/s.
(c) Evaluate x for vi = 135 m/s.
(d) Assume vi is close to but not equal to zero. Show that one term in the answer to part (a) dominates so that the function x(vi) reduces to a simpler form.
(e) If vi is very large, what is the approximate form of x(vi)?

(f) Describe the overall shape of the graph of the function x(vi).

Explanation / Answer

a. there is no acceleration in the x, so we just can make it vi*cos(36) (the component of initial velocity in the x. v*t=x. Now to find the time. Yf=0=1/2*a*t^2+v0*t+y0=1/2*-9.8*t^2+vi*sin(36)*t+2 solving the quadratic: t~0.0599781 (v+sqrt(v^2+113.462)) Plugging back in we get: 0.0599781 (v+sqrt(v^2+113.462))*v*cos(36)=x, b. plugging in 0.2 gets us .105 meters. c. plugging in 135 gets us 1771.42 d. if v initial is very small, then it looks very similar to a free fall problem. You should rework the problem without a v0*t in the y direction. e. If vi is very large, then the 113 in the square root really doesn't matter and it becomes (v+v)*v*cos(36)*0.056 f. it's a curve similar to the right half of a quadratic, with a horizontal asymptote in the negatives, going to infinity on the right side getting steeper and steeper

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