Question

A cannon-ball is fired horizontally off of the edge of a cliff. When the cannon-ball hits the water below it hits the water with a velocity of 19 m s−1 at an angle of 65° below the horizontal.

How tall was the cliff? (in m to 2.s.f)

(note: ignore the effects of air resistance when solving this problem)

Answer 1

Let us introduce a rectangular Cartesian coordinate system. $y$ - axis is pointing up, $y$ - coordinate of the top of the cliff is equal to its height $h$ . $y$ - coordinate of the bottom edge of the cliff is 0. $x$ -axis is pointing the same direction as the initial speed of the cannon-ball (see picture)

(thick brown line is the cliff)

Since we have to ignore  the effects of air resistance, the only force attracted to the cannon-ball is the gravity, which is directed vertically. The initial velocity $v_0$ is directed horizontally. So we can derive equations for axis projections of the velocity of the cannon-ball depending on time $t$ passed since the cannon-ball was thrown.

$$v_x(t)=v_0$$ $$v_y(t)=-gt$$

So we can derive equations for coordinates of the cannon-ball depending on time $t$ passed since the cannon-ball was thrown.

$$x(t)=v_0t$$

To derive analogical equation for $y$ we have to look at the graph of the equation $v_y=-gt$

The distance the cannon-ball has traveled along the $y$ - axis at the moment when the time $t$ has passed since the cannon-ball was thrown is equal to the area between the

graph of dependence of $v_y$ on time

and

the time-axis

This area is a rectangular triangle with side legs $gt$ and $t$. So its area is equal to $g*gt*\frac{1}{2}=\frac{gt^2}{2}$

Since the initial $y$ - coordinate of the cannon-ball is $h$ and the cannon-ball moves down, $y$ - coordinate of the cannon ball after the time $t$ will be

$$y(t)=h-\frac{gt^2}{2}$$

So we have 4 equations now:

$$v_x(t)=v_0$$ $$v_y(t)=-gt$$ $$x(t)=v_0t$$ $$y(t)=h-\frac{gt^2}{2}$$

The $T$ time passed from the moment when the cannon-ball was thrown to the moment when it hits the water.

$$y(T)=0 \rArr 0=h-\frac{gT^2}{2}\rArr T^2=\frac{2h}{g}$$

$v$ is the velocity of the cannon-ball at the moment when it hits the water

$v, v_x, v_y$ are sides of a rectangular triangle (see picture below)

Accirding to Pythagor's theorem,

$$v^2=v_x^2(T)+v_y^2(T)=v_0^2+g^2T^2=v_0^2+g^2\frac{2h}{g}=v_0^2+2gh$$

This picture presents relative position of vectors $v_y, v_x, v$ at the moment $T$

We know that the angle between the velocity and the horizon at this moment is $65^o$ .

$$tan (65^o)=\frac{v_y(t)}{v_x(t}\rArr tan^2(65^o)=\frac{v_y^2(t)}{v_x^2(t)}=\frac{2gh}{v_0^2}$$

$$v_0^2=\frac{2gh}{tan^2(65^o)}$$ $$v^2=v_0^2+2gh=\frac{2gh}{tan^2(65^o)}+2gh=2gh(1+\frac{1}{tan^2(65^o)})$$

$$v^2=2gh(\frac{1+tan^2(65^o)}{tan^2(65^o)})\rArr h=\frac{v^2(tan^2(65^o)}{2g(1+tan^2(65^o))}$$

$$h=\frac{19^2*tan^2(65^o)}{2*9,8*(1+tan^2(65^o))}=15.13 m$$