Let's call h the initial height of the rock (h=50 m, the height of the bridge).
Initially, the rock has only gravitational potential energy, which is given by
[tex]U_i=mgh[/tex]
where m=10 kg is the mass of the rock while g is the gravitational acceleration. So, the total mechanical energy of the rock at this point is
[tex]E_i = U_i = mgh=(10 kg)(9.81 m/s^2)(50 m)=4905 J[/tex]
At midway point of its fall, its height is [tex] \frac{h}{2} [/tex], so its potential energy is
[tex]U_f = mg \frac{h}{2} = (10 kg)(9.81 m/s^2)(25 m)=2452.5 J[/tex]
But now the rock is also moving by speed v, so it also has kinetic energy:
[tex]K_f = \frac{1}{2}mv^2 [/tex]
So the total energy at the midway point of the fall is
[tex]E_f = U_f + K_f[/tex] (1)
The mechanical energy must be conserved, so [tex]E_i = E_f[/tex], so we can rewrite (1) and solve it to find the kinetic energy of the rock at midway point of its fall:
[tex]E_i = U_f + K_f[/tex]
[tex]K_f = E_i - U_f = 4905 J - 2452.5 J=2452.5 J[/tex]
