Answer:139 ft/s
Explanation:
Given
ball leaves the bat at angle of [tex]75^{\circ} [/tex]
as the height with ball is launched and height of fielder is same so its vertical displacement is zero and ball is considered to complete a projectile motion with range =300 ft
Also range of projectile is given by
[tex]R=\frac{u^2sin2\theta }{g}[/tex]
where u= launch velocity
[tex]300=\frac{u^2sin150}{32.2}[/tex]
[tex]9660=u^2\times 0.5[/tex]
[tex]u^2=19,320[/tex]
[tex]u=\sqrt{19320}=138.99\approx 139 ft/s[/tex]
