(a) 23.1 m
The vertical velocity of the rock at time t is given by:
[tex]v_y(t) = v_{y0} + gt[/tex] (1)
where
[tex]v_{y0} = v_0 sin \theta = (33.0 m/s)(sin 25.3^{\circ})=14.1 m/s[/tex] is the initial vertical velocity of the rock
[tex]g=-9.8 m/s^2[/tex] is the acceleration due to gravity (negative because it is downward)
t is the time
At the point of maximum height, the vertical velocity is zero:
[tex]v_y(t)=0[/tex]
Using this information in eq.(1), we find the time it takes for the rock to reach the maximum height:
[tex]v_{y0} + gt=0\\t=-\frac{v_{y0}}{g}=-\frac{14.1 m/s}{-9.8 m/s^2}=1.44 s[/tex]
And now we can calculate the vertical position of the rock after t=1.44 s by using the equation:
[tex]y(t)=y_0 + v_{0y}t + \frac{1}{2}gt^2=13.0 m+(14.1 m/s)(1.44 s)+\frac{1}{2}(-9.8 m/s^2)(1.44 s)^2=23.1 m[/tex]
(2) 36.7 m/s
For this part, we have to calculate the time t at which the rock reaches the ground, which means y(t)=0. So:
[tex]y(t)=0=y_0 + v_{y0}t + \frac{1}{2}gt^2\\0=13.0 + 14.1t - 4.9t^2[/tex]
which has two solutions:
t = -0.73 s --> negative, we discard it
t = 3.61 s --> this is our solution
So now we can calculate the vertical velocity of the rock when it reaches the ground:
[tex]v_y(t)=v_{y0} + gt = 14.1 m/s+(-9.8 m/s^2)(3.61 s)=-21.4 m/s[/tex]
The horizontal velocity has not changed, since the motion along the horizontal direction is uniform, so it is
[tex]v_x(t)=v_{x0}=(33.0 m/s)(cos 25.3^{\circ})=29.8 m/s[/tex]
So, the magnitude of the velocity when the rock hits the ground is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(-21.4 m/s)^2+(29.8 m/s)^2}=36.7 m/s[/tex]
(3) 107.6 m
The horizontal distance travelled by the rock is given by:
[tex]d=v_x t[/tex]
where
[tex]v_x = 29.8 m/s[/tex] is the horizontal velocity, which is constant
[tex]t=3.61 s[/tex] is the time it takes for the rock to reach the ground
Substituting, we find
[tex]d=(29.8 m/s)(3.61 s)=107.6 m[/tex]
