Answer:
[tex]v_f = 31.44 m/s[/tex]
Explanation:
initial height of the ball is given as
[tex]H = 50.0 m[/tex]
initial speed of the ball is given as
[tex]v = 10.0 km/h[/tex]
now we know that
[tex]v = 10 \times \frac{1000}{3600}[/tex]
[tex]v = 2.78 m/s[/tex]
now by energy conservation we can say
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
[tex]\frac{1}{2}mv_i^2 + mgh_1 = \frac{1}{2}mv_f^2[/tex]
[tex]\frac{1}{2}(2.78)^2 + (9.81)(50.0) = \frac{1}{2}v_f^2[/tex]
[tex]v_f = 31.44 m/s[/tex]
