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:
[tex]S_x(t)=v_x t[/tex]
[tex]S_y(t)=h-v_y t- \frac{1}{2} g t^2 [/tex]
where h=654 m is the initial height of the projectile, [tex]v_x[/tex] and [tex]v_y[/tex] are the initial velocities on the x- and y-axis, g=9.81 m/s^2 and t is the time. [tex]v_y[/tex] ad [tex]g[/tex] have a negative signs because they both point downward.
We can calculate the initial velocity on the y-axis by requiring that [tex]S_y(6.70 s)=0[/tex], since we know that after 6.70 s the projectile reached the ground. Therefore:
[tex]0=h-v_yt- \frac{1}{2} gt^2=654 - 6.7 v_y - \frac{1}{2} (9.81)(6.70)^2 [/tex]
from which we find
[tex]v_y=64.78 m/s[/tex]
Then we can find the magnitude of the initial velocity v using the angle with respect to the vertical [tex]\alpha=40.9^{\circ}[/tex]:
[tex]v_y = v cos \alpha[/tex]
[tex]v= \frac{v_y}{cos \alpha} =85.7 m/s[/tex]
B) Let's calculate the component of the initial velocity on the x-axis:
[tex]v_x = v sin \alpha = 85.7 m/s \cdot sin (40.9^{\circ})=56.11 m/s[/tex]
And then, we can find how far the projectile traveled horizontally by calculating Sx at t=6.70 s, when it hits the ground:
[tex]S_x(6.70 s)=v_x t = 56.11 m/s \cdot 6.7 s=375.94 m[/tex]
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:
[tex]v_x (6.70 s)=v_x = 56.11 m/s[/tex]
D) Instead, vy changes during the motion since it's an accelerated motion, following the law
[tex]v_y(t) = - v_y -gt[/tex]
Using t=6.7 s, we can find the vertical velocity just before the projectile hits the ground
[tex]v_y(6.70 s)= -64.78 m/s - (9.81 m/s^2)(6.70 s)=-130.51 m/s[/tex]
Writing it with positive sign:
[tex]v_y(6.70 s) = 130.51 m/s[/tex]
