The maximum horizontal distance traveled by the base ball is the horizontal distance it covers when it returns to its initial height. Let it be # (y = 0)# as shown below.

If the ball is thrown making an angle #theta# with the horizontal with velocity #v_0#

Using the kinematic equation

#s=ut+1/2g t^2#,

Noting that acceleration due to gravity #g=9.81ms^-2# is opposing the direction of motion. we get for vertical component of velocity

#0=v_{0}t\sintheta -\frac {1}{2}g t^{2}#

#=>0=v_{0}\sintheta -\frac {1}{2}g t#

Time #t# taken to return to ground or to travel distance #d# is found as

#t = (2 v_0 sin θ) / g# ....(1)

Ignoring air resistance,

Distance #d="Horizontal component of velocity"xx"time"#

#=(v_{0}cos\theta )t# .....(2)

Inserting value of #t# from (1) in (2) we get

#d=(v_{0}cos\theta )(2 v_0 sin θ) / g#

#=>d=v_{0}^2/g 2sin theta cos\theta #

#=>d=v_{0}^2/g sin 2theta #

For its maximum range #sin 2 theta=1#

#=>d_max=v_{0}^2/g #

Taking Inserting values we get

#d_max=(35.0)^2/9.81 #

#=>d_max=124.87m#, rounded to two decimal places.