A cannon is positioned with a direction of 60 degrees between the ground and itself. (Sorry, again, for my poor English. I hope you understood that sentence.). The shooting velocity is $800{\text{m}\over \sec}$. What is its range?
I didn't really understand this question. Am I asked how further the cannon ball will go before it descends? Because if so, I should I find the highest point the cannon ball get? Should I do it by differentiating?
I would truly appreciate your assistance.
On
In projectile motion, you need to concern yourself with the two components of displacement:
$$x=x_0 + v_x t + \frac{1}{2}a_x t^2$$ $$y=y_0 + v_y t + \frac{1}{2}a_y t^2$$
If we place the cannon at the origin, then $(x_0,y_0) = (0,0)$. Furthermore, there is no acceleration in the $x$ direction if we neglect things like air resistance. Then $a_x = 0$. The acceleration in the $y$ direction is due to gravity $a_y = -g$.
$$x=v_x t$$ $$y=v_y t - \frac{1}{2}g t^2$$
We need to find the range. That is, we need to find the displacement in $x$ that occurs when the cannonball hits the ground, or when $y=0$.
$$x=v_x t$$ $$0=v_y t - \frac{1}{2}g t^2$$
The velocity can be broken up into components with trigonometry.
$$v_x = 800\cos60^\circ m/s$$ $$v_y = 800\sin60^\circ m/s$$
Using this information, solve for $t$ when the cannonball hits the ground. Then use that time to calculate displacement in $x$.
In the simplest model, you are working in two dimensions, with the system of equations:
$$\frac{d^2 y}{dt^2} = -g \\ \frac{d^2 x}{dt^2} = 0 \\ y(0)=0,y'(0)=\sin(\theta) s \\ x(0)=0,x'(0)=\cos(\theta) s $$
where $g>0$ is acceleration due to gravity, $\theta$ is the angle of the initial velocity, and $s$ is the initial speed. Note that this model assumes no friction. When you solve you find
$$y(t)=-\frac{1}{2} gt^2 + \sin(\theta) s t \\ x(t)=\cos(\theta) s t.$$
Now the range of the ball is the horizontal distance it travels before hitting the ground, i.e. the value of $x$ when $y=0$. Can you find this using these equations?