The initial velocity of the ball is 0. Applying:
v = u + at
v = 0 + 229 x 0.08
v = 18.3 m/s
a)
Vx = Vcos(∅)
Vx = 18.3cos(52.3)
Vx = 11.2 m/s
b)
Vy = Vsin(∅)
Vy = 18.3sin(52.3)
Vy = 14.5 m/s
PART A)
Here by force balance along Y direction

Force balance along X direction

now by torque balance



PART B)
now from above equations



now net reaction force of wall is given as


Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.