Answer:
19.76 m/s
Explanation:
Let v be the speed just before the ball hits the ground. The v/3 is the initial speed. When the ball falls to the ground its potential energy is converted to kinetic energy: [tex]E_p + E_{k0} = E_k[/tex][tex]mgh + mv_0^2/2 = mv^2/2[/tex]where m is the mass and h = 17.7 is the vertical distance traveled, [tex]v_0 = v/3[/tex] is the initial velocity , and g = 9.81 m/s2 is the gravitational acceleration. We can divide both sides by m:
[tex]gh + v^2/18 = v^2/2[/tex]
[tex]gh = 9v^2/18 – v^2/18 = 8v^2/18 = 4v^2/9[/tex]
[tex]9.81*17.7 = 4v^2/9[/tex]
[tex]v^2 = 390.68[/tex]
[tex]v = \sqrt{390.68} = 19.76m/s[/tex]
