Question

an underground cannon launches a cannonball from ground level at a 35-degree angle. the cannonball is shot with an initial velocity of 15 m/s.

Answer 1

Answer:

$\displaystyle x_{max}=21.57\ m$

$\displaystyle y_{max}=3.78\ m$

Explanation:

Motion in Two-Dimensions

When an object is launched with a certain angle $\theta$ above ground level with an initial velocity $\vec v_o$, it describes a curve called a parabola, defined by the force of gravity which will eventually make the object return to the ground. There are two overlapping motions, the horizontal motion, at a constant speed, and the vertical motion, at changing speed because the acceleration of gravity modifies it. The velocity $\vec v_o$ is split into two components x,y

$\displaystyle v_{ox}=|vo|\ Cos\theta$

$\displaystyle v_{oy}=|vo|\ Sin\theta$

The position of the object is also split into its components, assuming the object was launched from ground level

$\displaystyle x=v_{ox}.t$

$\displaystyle y=v_{oy}.t-\frac{g.t^2}{2}$

The maximum horizontal distance the object reaches (called range) is

$\displaystyle x_{max}=\frac{2v_{ox}\ v_{oy}}{g}$

The maximum height is given by

$\displaystyle y_{max}=\frac{v_{oy}^2}{2g}$

The question doesn't ask for anything in particular, but to guide you in the solution of your own problem, we'll compute $X_{max}$ and $Y_{max}$ for you. The data is

$\displaystyle |vo|=15\ m/s\ ,\ \theta =35^o$

Let's compute the range

$\displaystyle x_{max}=\frac{(2)(15)\ cos35^o\ (15)\ sen35^o}{9.8}$

$\displaystyle x_{max}=\frac{211.43}{9.8}=21.57\ m$

Now for the maximum height

$\displaystyle y_{max}=\frac{(15\ . \ Sin35^o)^2}{2(9.8)}$

$\displaystyle y_{max}=\frac{74.0227}{19.6}$

$\displaystyle y_{max}=3.78\ m$