Gypsy Danger is a behemoth, weighing in at 7,080 tons. How much of an impact would it cause, when it fell from the sky in Pacific Rim: Uprising? Would it be enough to kill the Mega-Kaiju? Challenge accepted.
Lets assume that the jetpacks used for the Jaegers are just as powerful as NASA's Saturn V rocket. The "jetpack" alone would weigh around 480,000 kg. Saturn V rockets produce a force of 4,900,000 N. Gypsy danger itself weighs 7,080,000 kg.
Only SI units will be used in all calculations.
$$F=ma,\; \therefore a={{F}\over{m}}$$ $$\therefore a={4,900,000 \over 480,000+7,080,000}={4,900,000 \over 7,560,000}=0.65\;ms^{-2}$$ The rocket carries both its own weight, and also the weight of the Jaeger. Hence both their weights are added.
This acceleration is incredibly slow, but it's okay since we're dragging a giant robot along with us. There is also a lot of crashing into buildings, which we will conveniently ignore to make our calculations easier.
Analyzing the above clip, we see that the jetpack is on for a total of 45 seconds, give or take. In 45 seconds, Gypsy Danger along with the rocket would accelerate a lot:
$$a={v \over t},\; \therefore v=a×t$$ $$\therefore v=0.65×45=29.25\; ms^{-1}$$ $$\therefore {S\over t}=29.25,\; \therefore S=29.25×45$$ $$\therefore S=1316.25 \;m$$
See the Math
In the above picture, Gypsy Danger is seen heading upwards at a roughly 45° angle. We can now find out the vertical height of Gypsy Danger just before falling down.
In the above graph, O is the starting point. Y-axis is the height above ground, and the X-axis is the horizontal distance. Line A shows the upward trajectory of the Jaeger.
Both d1s are the same, since any right triangle with 45° is isosceles. Taking \(A\) as the hypotenuse of the triangle, which is 1316.25 m as shown before this section. By the pythagoras theorem: $$A^2={d_1}^2+{d_1}^2=2×{d_1}^2$$ $$\therefore d_1=\sqrt{A^2\over 2}=\sqrt{1316.25×1316.25\over 2}=930.73\; m$$
Doing the calculations, we find that the Jaeger is at a maximum height of about 930.73 metres before falling down on top of the Mega-kaiju.
Now, we calculate the final speed of the Jaeger just before reaching the ground. Using the equations for free fall in air, we can find \(v_f\). We can completely ignore the terminal velocity since the Jaeger is so heavy.
$$\text{In free fall,}\;\; h = {gt^2\over 2}$$ $$\therefore t=\sqrt{h×2\over g}=\sqrt{930.73×2\over 9.8}=13.77\;s$$
$$\text{Now,}\;\; v_f=g×t$$ $$\therefore v_f=9.8×13.77≈135\; ms^{-1}$$
On a side note, Jake and Amara only have about 10 seconds to escape the Jaeger before it hits the ground. That's a really short span of time, and it also doesn't match up with the timing in the movie.
Thus, the Jaeger is travelling at approximately \(135\; ms^{-1}\) just before it hits the Mega-kaiju. Let's calculate the kinetic energy of the Jaeger just before it hits:
$$\text{K.E.}={{m×v^2}\over 2}$$ $$\therefore \text{K.E.}={{1,070,000×135^2}\over 2}=9,750,375,000 \;\text J≈ 0.01\;\text {terajoules or TJ}$$
0.01 Terajoules is a pretty small number. In contrast, the atom bomb detonated at Hiroshima during World War 2 released 63 terajoules of energy.
It's still okay, though, since 0.01 Terajoules is also the energy released by explosion of 1 kiloton of TNT. But the Mega-kaiju would most likely shrug it off easily. See any calculation or logical errors? Comment down below.
In the end, of course, it all depends on the film makers. Because who cares about science when you've got seven million kilograms of steel all revved up and ready to fight?