F=ma: mg+kv=mdv/dt and initial condition V(t=0)=60m/s, V=f(t) can be obtained.

When the object reaches the maximum height, V=f(t)=0, and the time t can be solved.

The maximum height Vdt can be multiplied by a definite integral from 0 to t, which is the height.

The solution of the above differential equation is V=(m/k)(exp(kt-c)-g), where c is a constant. By introducing initial conditions, the following problems can be solved.