(A) 28.1 m
The initial velocities of the rock along the x (horizontal) and y (vertical) directions are
![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](https://tex.z-dn.net/?f=v_x%20%3D%20%2830.0%20m%2Fs%29%20cos%2034.9%5E%7B%5Ccirc%7D%3D24.6%20m%2Fs%5C%5Cv_y%20%3D%20%2830.0%20m%2Fs%29%20sin%2034.9%5E%7B%5Ccirc%7D%20%3D17.2%20m%2Fs)
The vertical velocity of the rock at time t is given by
![v(t) = v_y -gt](https://tex.z-dn.net/?f=v%28t%29%20%3D%20v_y%20-gt)
where
is the initial vertical velocity and
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:
![0=v_y -gt\\t=\frac{v_y}{g}=\frac{17.2 m/s}{9.8 m/s^2}=1.76 s](https://tex.z-dn.net/?f=0%3Dv_y%20-gt%5C%5Ct%3D%5Cfrac%7Bv_y%7D%7Bg%7D%3D%5Cfrac%7B17.2%20m%2Fs%7D%7B9.8%20m%2Fs%5E2%7D%3D1.76%20s)
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
![y(t) = y_0 + v_y t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_0%20%2B%20v_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
where
is the initial height. Substituting t=1.76 s, we find
![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](https://tex.z-dn.net/?f=y_%7Bmax%7D%3D13.0%20m%20%2B%20%2817.2%20m%2Fs%29%281.76%20s%29-%5Cfrac%7B1%7D%7B2%7D%289.8%20m%2Fs%5E2%29%281.76%20s%29%5E2%3D28.1%20m)
(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:
![0=y_0 + v_y t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=0%3Dy_0%20%2B%20v_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
Substituting numbers, it becomes
![0=13+17.2 t -4.9t^2](https://tex.z-dn.net/?f=0%3D13%2B17.2%20t%20-4.9t%5E2)
which gives two solutions:
(negative, so physically meaningless: we discard it)
--> 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:
![v_y(t)=v_0 -gt=17.2 m/s-(9.8 m/s^2)(4.15 s)=-23.5 m/s](https://tex.z-dn.net/?f=v_y%28t%29%3Dv_0%20-gt%3D17.2%20m%2Fs-%289.8%20m%2Fs%5E2%29%284.15%20s%29%3D-23.5%20m%2Fs)
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
, the magnitude of the resultant velocity is
![v=\sqrt{v_x^2+v_y^2}=\sqrt{(24.6 m/s)^2+(-23.5 m/s)^2}=34.0 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D%3D%5Csqrt%7B%2824.6%20m%2Fs%29%5E2%2B%28-23.5%20m%2Fs%29%5E2%7D%3D34.0%20m%2Fs)
(C) 102.1 m
Since the rock is moving by uniform motion along the x-axis, the horizontal distance is simply given by:
![d_x = v_x t](https://tex.z-dn.net/?f=d_x%20%3D%20v_x%20t)
and substituting the total time of the fall, t=4.15 s, we find
![d_x = (24.6 m/s)(4.15 s)=102.1 m](https://tex.z-dn.net/?f=d_x%20%3D%20%2824.6%20m%2Fs%29%284.15%20s%29%3D102.1%20m)