Complex SUVAT help? Two objects are dropped from the top of a cliff height $H.$ The second is dropped when the first has travelled a distance $D.$ Prove that the instant when the first object has reached bottom, the second object is at a distance above the ground $2\sqrt{DH} - D$.
What is the process of completing this question? I have been trying for hours and nothing seems to be getting me even remotely close to the answer. I understand that once I have figured out the distance $D,$ it is a case of subtracting it from $H$ and rearranging... but nothing is working.
thanks :)
 A: Time taken for object 1 to travel distance D: we use $s = ut + \frac{1}{2}at^2 \ $ to get $\ t_D = \sqrt{\frac{2}{g}} \cdot \sqrt{D}$.
Time taken for object 1 to get to the ground: we use $s = ut + \frac{1}{2}at^2 \ $ again to get $\ t_H = \sqrt{\frac{2}{g}} \cdot \sqrt{H}$.
So the time between object 2 being released and object 1 hitting the ground is $\ T = \sqrt{\frac{2}{g}}\bigg(\sqrt{D} - \sqrt{H}\bigg)$.
The distance travelled by object 2 in time T: we use $s = ut + \frac{1}{2}at^2 \ $ again to get $$s = \frac{g}{2}\bigg(\sqrt{\frac{2}{g}}\big(\sqrt{H} - \sqrt{D}\big) \bigg)^2 = H + D -2\sqrt{D \cdot H}.$$
Therefore, object 2 is a distance $$H - (H + D -2\sqrt{D \cdot H}) = 2\sqrt{D \cdot H} -D$$ off the ground when object 1 hits the ground.
A: If $a$ is the acceleration due to gravity, then the distance during time $t$ after the object is dropped from rest is $$\frac12at^2.$$
If $t_1$ is the time it takes for the object to travel distance $D$ then $$D=\frac12at_1^2\implies t_1=\sqrt{2D\over a}$$ 
Likewise, if $t_2$ is the time it takes the object to strike the ground, we have $$H=\frac12at_1^2\implies t_2=\sqrt{2H\over a}$$
Then distance the second object travels between times $t_1$ and $t_2$ is $$\frac12a(t_2-t_1)^2$$
Can you take it from here?
