107 139
Assignments Done
100%
Successfully Done
In January 2023

# Answer to Question #105934 in Mechanics | Relativity for Matthew

Question #105934
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)
1
2020-03-19T11:39:27-0400

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\n0=h-\\frac{gT^2}{2}\\rArr\nT^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"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!