Let's solve the problem step-by-step.

A) Let's start writing the laws of motion on both x (horizontal) and y (vertical) axis. On the x-axis, it's a uniform motion with constant velocity, while on the y-axis it's an uniformly accelerated motion:

where h=654 m is the initial height of the projectile,

and

are the initial velocities on the x- and y-axis, g=9.81 m/s^2 and t is the time.

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have a negative signs because they both point downward.

We can calculate the initial velocity on the y-axis by requiring that

, since we know that after 6.70 s the projectile reached the ground. Therefore:

from which we find

Then we can find the magnitude of the initial velocity v using the angle with respect to the vertical

:

B) Let's calculate the component of the initial velocity on the x-axis:

And then, we can find how far the projectile traveled horizontally by calculating Sx at t=6.70 s, when it hits the ground:

C) The horizontal component of the velocity does not change during the motion, since it's an uniform motion on the x-axis. Therefore, vx at t=6.7 s is the same as its initial value:

D) Instead, vy changes during the motion since it's an accelerated motion, following the law

Using t=6.7 s, we can find the vertical velocity just before the projectile hits the ground

Writing it with positive sign: