Ch4 #57 A spring cannon is located at the edge of a table that is 1.20 m above t
ID: 2164059 • Letter: C
Question
Ch4 #57
A spring cannon is located at the edge of a table that is 1.20 m above the floor. A steel ball is launched from the cannon with speed vi at 35.0: above the horizontal, 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). Evaluate x for vi = 0.100 m/s and for vi = 100 m/s. 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, If vi is very large, what is the approximate form of x(vi)? Describe the overall shape of the graph of the function x(vi).Explanation / Answer
Consider the y component of the problem first: vy0 = v0*sin(35) and it follows that the distance travelled, y, is y =y0+vy0*t - 0.5*g*t^2 y0 = 1.2 so y = 1.2+vy0*t - 0.5*g*t^2 The ball hits the floor when y=0 So, you now have a quadratic... 0.5*g*t^2-vy0*t-1.2 = 0 Solve t = (vy0 + sqrt(vy0^2 + 4*0.5g*1.2))/g (the answer has to be the + root) t = (vy0 + sqrt(vy0^2 + 2.4g))/g You now have the time it takes to hit the floor. Now look at the x direction... vx0 = v0*cos(35) x = x0 +vx0*t (no acceleration in the x-direction) So, x= v0*cos(35)*(v0*sin(35) + sqrt((v0*sin(35))^2 + 2.4g))/g after all the substitutions