a) 5.6 m/s
First of all, we consider the vertical motion of the stone, which is a free fall motion (uniform accelerated motion). We can use the suvat equation :
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 200 m is the vertical displacement (here we chose downward as positive direction)
u = 0 is the initial vertical velocity of the stone (since it is launched horizontally)
t is the time
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find the time the stone takes to reach the ground:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(200)}{9.8}}=6.39 s[/tex]
Now we can consider the horizontal motion: this is a uniform motion with constant speed. The horizontal distance travelled is given by
[tex]d=v_x t[/tex]
where
d = 36.0 m is the horizontal distance travelled by the stone
t = 6.39 s is the time of flight
If we solve for [tex]v_x[/tex], we find the speed at which the stone was launched (which remains constant during the whole motion):
[tex]v_x = \frac{d}{t}=\frac{36}{6.39}=5.6 m/s[/tex]
(b) 62.8 m/s at [tex]84.9^{\circ}[/tex] below the horizontal
- The motion along the horizontal direction is a uniform motion, since there are no forces acting on the stone in this direction - so the horizontal velocity remains constant:
[tex]v_x = 5.6 m/s[/tex]
- The vertical velocity instead changes due to the effect of the acceleration of gravity. We can calculate the vertical velocity at the time of impact by using the equation
[tex]v_y = u_y + gt[/tex]
where
[tex]u_y = 0[/tex] is the initial vertical velocity (zero because the stone is launched horizontally)
Solving for t = 6.39 s, we find:
[tex]v_y = 0+(9.8)(6.39)=62.6 m/s[/tex]
Keep in mind that the direction of this velocity is downward.
So now we can find the speed of the stone at the moment of impact:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{5.6^2+62.6^2}=62.8 m/s[/tex]
And the angle of impact, measured as below the horizontal (since the vertical velocity is downward) is
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{62.6}{5.6})=84.9^{\circ}[/tex]
