How to obtain the distance covered by a projectile if an additional acceleration is taken into account? The problem is as follows:
A projectile is fired from a horizontal ground with a speed of $\vec{v}=10\hat{i}+40\hat{j}\,\frac{m}{s}$. Due the gravity and the wind the shell experiences a constant acceleration of $\vec{a}=-5\left(\hat{i}+2\hat{j}\right)\frac{m}{s^{2}}$. Find in meters from what distance from the point the projectile was fired the shell hit the ground.
$\begin{array}{ll}
1.&40\,m\\
2.&80\,m\\
3.&120\,m\\
4.&100\,m\\
\end{array}$
I'm not very sure of how to use the vectors given to find the distance. Typically I would use the equation of the position for the horizontal range would be:
$x(t)=x_{o}+v_{o}\cos\omega t$
However since it mentions that there's an acceleration that is produced by the wind and gravity. How would I use this information.
Would it mean that should I take the norm for that acceleration as follows?
$\left \| \left \langle -5,-10 \right \rangle \right \|=\sqrt{125}=5\sqrt{5}$
and from that should I use the acceleration instead $g=9.8$, $g=5\sqrt{5}$ as in the equation from below?
$y(t)=y_{o}+v_{o}\sin\omega t - \frac{1}{2}at^{2}$
Since it is fired from the ground would it mean:
$y(t)= v_{o}\sin\omega t - \frac{1}{2}at^{2}$ ?
The only information I could obtain was:
$\sin\omega=\frac{40}{\sqrt{1700}}=\frac{4}{\sqrt{17}}$
But that's how far I went with this problem. Can somebody offer me some help with this?. Supposedly the answer is $80$ but I don't know what to do to get there.
 A: When the projectile hits the ground, its velocity is -40m/s, the reverse of the initial velocity,. The time of flight, $t$, is given by the deceleration formula along with the deceleration $-10m/s^2$,
$$(-40)-40=-10t$$
which yields $t=8s$. Then, with the initial horizontal velocity $10m/s$ and the deceleration $-5m/s^2$, the horizontal distance travelled is
$$ d= 10t-\frac 12(5)  t^2= 80 - 160 = -80m$$
Thus, the answer is (2). Note that the wind reverses the horizontal travel direction of the projectile and it lends 80m on the opposite side.
A: $/ \vec{i}$
$-5 = ax$
Then $x= \int \int ax dt = -\frac{5}{2}t^2+10*t+x0$
$/ \vec{j}$
$-10=ay$
Then $y= -5t^2+40*t+y0$
Assuming $y0=0$, $y=-5t^2+40t$.
$y(t)=0$ for $t=${$0,8$} 
$x(t=8)= -80m$ (with x0=0), also I’m not sure about the direction of the vectors.
A: The problem is wrongly spelled.
If i is the horizontal direction and j vertical, then the horizontal travel is negative, -80m. In this case you can imagine that for the amount of time it was above the ground, the projectile travelled against the wind and ended up being blown 80m back past the launch point. This is definitely confusing but is the only way to interpret the 80m answer listed.
With i vertical and j horizontal, the equations of motion in scalar form are:
$x=40t-10\frac{t^2}{2}$ 
$y=10t-5\frac{t^2}{2}$
then y=0 solves for t, t=2 which solves for x, x=60m. This answer is not listed
