Answer:139 ft/s
Step-by-step explanation:
Given
ball leaves the bat at angle of
![75^(\circ)](https://img.qammunity.org/2020/formulas/physics/college/eyqonpyg8rzg83jav2zp8rcv7qs4ktg03r.png)
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
![R=(u^2sin2\theta )/(g)](https://img.qammunity.org/2020/formulas/physics/college/u67f043ld4trt01h4npk7bi5vdlcjt3ymq.png)
where u= launch velocity
![300=(u^2sin150)/(32.2)](https://img.qammunity.org/2020/formulas/physics/high-school/c944howvq7k14o3khpvgff5j01kiwv5zjj.png)
![9660=u^2* 0.5](https://img.qammunity.org/2020/formulas/physics/high-school/n5vhpisr78kc82upkq74jfz99hthaxj2yp.png)
![u^2=19,320](https://img.qammunity.org/2020/formulas/physics/high-school/t7pl6wvlwp58fxqiezuccn5zn1sy1wusfs.png)
![u=√(19320)=138.99\approx 139 ft/s](https://img.qammunity.org/2020/formulas/physics/high-school/k38e9cbs7zafqkdi3jwt8vmg91uirhngcy.png)