Refer to the diagram show below.
Let V = the speed of the ball when it left the bat.
The horizontal component of velocity is
Vx = Vcos(45°) = 0.7071V
The vertical component of velocity is
Vy = V sin(45°) = 0.7071V
Assume g = 9.8 m/s² and ignore air resistance.
The time of flight is
t = (80 m)/(Vx m/s) = 80/(0.7071V) = 113.1382/V s
The difference in height between the launch point and the fence is
h = 10 - 0.9 = 9.1 m.
Therefore
[tex](0.7071V \, \frac{m}{s})( \frac{113.1382}{V} \, s) - \frac{1}{2}(9.8 \, \frac{m}{s^{2}} ) ( \frac{113.1382}{V} \, s)^{2} = (9.1 \, m) [/tex]
[tex]80- \frac{6.2721 \times 10^{4}}{V^{2}} =9.1 \\\\ V^{2} = \frac{6.2721 \times 10^{4}}{70.9} \\\\ V = 29.743 \, \frac{m}{s} [/tex]
Answer: 29.7 m/s (nearest tenth)
