(A) 28.1 m
The initial velocities of the rock along the x (horizontal) and y (vertical) directions are
[tex]v_x = (30.0 m/s) cos 34.9^{\circ}=24.6 m/s\\v_y = (30.0 m/s) sin 34.9^{\circ} =17.2 m/s[/tex]
The vertical velocity of the rock at time t is given by
[tex]v(t) = v_y -gt[/tex]
where [tex]v_y = 17.2 m/s[/tex] is the initial vertical velocity and [tex]g=9.8 m/s^2[/tex] is the gravitational acceleration.
At the point of maximum height, the vertical velocity is zero: v(t)=0, so we can calculate the time t at which this occurs:
[tex]0=v_y -gt\\t=\frac{v_y}{g}=\frac{17.2 m/s}{9.8 m/s^2}=1.76 s[/tex]
So, the rock has reached its maximum height after t=1.76 s. Now we can calculate its maximum height with the equation for the vertical position
[tex]y(t) = y_0 + v_y t - \frac{1}{2}gt^2[/tex]
where [tex]y_0 = 13.0 m[/tex] is the initial height. Substituting t=1.76 s, we find
[tex]y_{max}=13.0 m + (17.2 m/s)(1.76 s)-\frac{1}{2}(9.8 m/s^2)(1.76 s)^2=28.1 m[/tex]
(B) 34.0 m/s
We need to find the time at which the rock hits the ground. We can do it by requiring y(t)=0 in the equation of the vertical position, so:
[tex]0=y_0 + v_y t - \frac{1}{2}gt^2[/tex]
Substituting numbers, it becomes
[tex]0=13+17.2 t -4.9t^2[/tex]
which gives two solutions:
[tex]t=-0.64 s[/tex] (negative, so physically meaningless: we discard it)
[tex]t=4.15 s[/tex] --> this is our solution, the time at which the rock hits the ground
Now we can substitute t=4.15 s in the equation of the vertical velocity, to find the vertical velocity of the rock as it strikes the ground:
[tex]v_y(t)=v_0 -gt=17.2 m/s-(9.8 m/s^2)(4.15 s)=-23.5 m/s[/tex]
The negative sign only means the direction is downward. However, this is only the vertical component of the velocity: since the rock is also moving along the horizontal direction, with constant velocity [tex]v_x[/tex], the magnitude of the resultant velocity is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(24.6 m/s)^2+(-23.5 m/s)^2}=34.0 m/s[/tex]
(C) 102.1 m
Since the rock is moving by uniform motion along the x-axis, the horizontal distance is simply given by:
[tex]d_x = v_x t[/tex]
and substituting the total time of the fall, t=4.15 s, we find
[tex]d_x = (24.6 m/s)(4.15 s)=102.1 m[/tex]
