Let us draw upon the fact that the horizontal and vertical components of a body that is falling are independent of each other. This is also known as Galileo's law of falling bodies. Now, let us assume that gravity is constant everywhere and a fall from a distance d that is solely dependent on the effect of gravity. Therefore we form the equation,\[a= 32.2 = constant\]Now, we know from mechanics that acceleration is the rate of change of velocity therefore we can put write this mathematically as such,\[a = \frac{dv}{dt}\]If we multiply by dt, and integrate the equation we get,\[\int\limits_{0}^{t}adt=\int\limits_{v_0}^{v}dv\]The result of this integration is this one, and this is one of the easiest integrals in mathematics.\[v-v_0 = at\]We put v0 on the RHS and we get,\[v=v_0+at\]Now we know that velocity is the rate of change of the distance, so we make this equation:\[v = \frac{ds}{dt}\]Let s=distance. Now multiply by dt, and input the v on the equation.\[(v_0+at)dt=ds\]Integrate,\[\int\limits_{0}^{s}ds=\int\limits_{0}^{t}v_0dt+\int\limits_{0}^{t}atdt\]This gives you,\[s=v_0t+\frac{1}{2}at^2\]We know that your case deals with zero initial velocity, so set v0=0. And we know that s=1000. But we want to solve for t(time).\[s=\frac{1}{2}at^2\]Multiply by 2 and divide by a.\[t^2=\frac{2s}{a}\]Take the square root and you get the correct equation.\[t=\sqrt{\frac{2s}{a}}\]Keeping in mind that a is constant and is equal to 32.2 and s is equal to 1000, we input hte numeric values and get.\[t=\sqrt{\frac{2*1000}{32.2}}=\sqrt{\frac{2000}{32.2}}=\sqrt{62.1118}=7.8811\]