Physics

Assuming the 75.0 J of work on the rock does not change its speed, the initial velocity of the rock is given by $v_i = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2 \times 75.0 J}{1.60 kg}} = 5.77 {\rm m/s}$ The rock will reach a maximum height when it comes to a stop, which is when its kinetic energy has been converted to gravitational potential energy. $mgh = \frac{1}{2}mv_i^2$ $h = \frac{v_i^2}{2g} = \frac{(5.77 {\rm m/s})^2}{2 \times 9.81 {\rm m/s^2}} = 29.4 {\rm m}$ The rock will go up to a height of 29.4 m.