Answer:
[tex]v_{o} = 79.314\times 10^{6}\,\frac{m}{s}[/tex]
Explanation:
Let assume that John Lester has a height of 1.80 meters and throws the ball at 70 percent of John Lester's height. The time before the ball hits the soil is:
[tex]0\,m = 0.7\cdot (1.80\,m) -\frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
[tex]t \approx 0.507\,s[/tex]
The initial horizontal velocity required to pitch the ball all the way around the Earth is:
[tex]2\pi\cdot (6.4\times 10^{6}\,m)= v_{o}\cdot (0.507\,s)[/tex]
[tex]v_{o} = 79.314\times 10^{6}\,\frac{m}{s}[/tex]
