A ladder 25ft long leans against a vertical wall with its foot on level ground 7 ft from the base of the wall. If the foot is pulled away from the wall at the rate 2 ft/sec, express the distance (y ft) of the top of the ladder above the ground as a function of the time t sec in moving.

Ans: y= 2(sqrt(144 - 7t - t^2)

Can someone explain this one to me?