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

`\displaystyle x_(max)=21.57\ m`

`\displaystyle y_(max)=3.78\ m`

Step-by-step 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-(g.t^2)/(2)`

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

`\displaystyle x_(max)=(2v_(ox)\ v_(oy))/(g)`

The maximum height is given by

`\displaystyle y_(max)=(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)=((2)(15)\ cos35^o\ (15)\ sen35^o)/(9.8)`

`\displaystyle x_(max)=(211.43)/(9.8)=21.57\ m`

Now for the maximum height

`\displaystyle y_(max)=((15\ . \ Sin35^o)^2)/(2(9.8))`

`\displaystyle y_(max)=(74.0227)/(19.6)`

`\displaystyle y_(max)=3.78\ m`