A spring cannon is located at the edge of a table that is 1.2m above the floor.

Question

A spring cannon is located at the edge of a table that is 1.2m above the floor. A steel ball is launched from the cannon withspeed Vo at 35.0 degrees above the horizontal.
(a) Find the horizontal displacement component of the ball tothe point where it lands on the floor as a function of Vo. We writethis function as x(Vo). Evaluate x for (b) Vo = 0.100 m/s and for(c) Vo = m/s. (d) Assume Vo is close to zero but not equal to zero.Show that one term in the answer to part (a) dominates so that thefunction x(Vo) reduces to a simpler form. (e) If Vo is ver large,what is the approximate form of x(Vo
I found the answer to a b and c but i am stuck on d and e.

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 That's it.

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