A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is $40 i + 48 k$, with speed measured in feet per second. The spin of the ball results in a southward acceleration of 6 $ft/s^2$, so the acceleration vector is $a = −6 j − 32 k$.
Where does the ball land? (Round your answers to one decimal place.) _______ ft from the origin at an angle of _____ ° from the eastern direction toward the south.
So I was able to find that the ball lands 123 ft from the origin based on the position function, but I'm unsure of how to find the angle. I can't really imagine the scenario, which I think is the main problem.
The position function that I found by the way is $r(t) = <40t, -3t^2, 48t-16t^2>$ with the ball landing at 3 sec.
I achieve the same general position vector as you do, but I'll work through the steps here anyway for any onlookers.
We know that $\mathbf{r}(0)=\mathbf{0},\mathbf{r}'(0)=\begin{bmatrix}40 \\0 \\ 48\end{bmatrix}$ which are our initial conditions and that our acceleration is constant throughout; given by $\mathbf{r}''(t)=\begin{bmatrix}0 \\-6 \\ -32\end{bmatrix}$.
Now we can find the particular integral to denote velocity (rate of change of position) using our expression for the acceleration and the initial value for velocity.
$$\mathbf{r}'(t)=\int{\mathbf{r}''(t)\, dt}=\int{\begin{bmatrix}0 \\-6 \\ -32\end{bmatrix}\,dt}=t\begin{bmatrix}0 \\-6 \\ -32\end{bmatrix}+\mathbf{c}$$
Using our initial condition for velocity we yield:
$$\mathbf{r}'(t)=t\begin{bmatrix}0 \\-6 \\ -32\end{bmatrix}+\begin{bmatrix}40 \\0 \\ 48\end{bmatrix}$$
This vector equation is the key to the question.
It is important to always visualise the velocity - at an instant - having the direction tangential to the particle path (i.e. graph of $\mathbf{r}(t)$) at that moment in time.
Therefore, when the particle hits the ground (i.e. when $z=0\mathbf{k}$) we know that the direction of the particle at that instant (say $t_c$ represents the time at which the particle hits the ground) is equal to the velocity vector at $t_c$ - denoted by $\mathbf{v}(t_c) = \mathbf{r}'(t_c)$.
Having calculated this velocity vector at $t_c$, we can hence calculate the angle of collision by making use of the dot product of two vectors:
$$\mathbf{v} . \mathbf{w}=||\mathbf{v}||.||\mathbf{w}||\cos{\theta}$$.
Remember that to find the correct angle (I assume the acute angle), you must have both the ground vector and the velocity vector moving towards their intersection.