Question

A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/s at angle 30° above the horizontal and hits the level ground at a horizontal distance 100m from the wall.

a) Calculate the time it takes the cannon ball to hit the ground.
b) Calculate the height of the castle wall.
c) What are the x- and y-components of the cannon ball’s velocity at the highest point of its trajectory?
d) What are the x- and y-components of the cannon ball’s velocity just before it hits the ground?
e) Sketch, qualitatively, x-t, y-t, vx-t and vy-t graphs for the cannon ball’s motion.

Answer 1

Step-by-step explanation:

Let's establish our knowable variables

`v_(o)=30 (m)/(s)`, ∅=30º

`x=100m`

a) Since we know the x distance that the cannonball hits the ground and the velocity at the x-axis is constant, meaning that the acceleration at x is 0.

We can calculate the time it takes the cannonball to hit the ground by using the following formula:

`x=v_(x)*t`

`t=(x)/(v_(x) )=(100m)/(30(m)/(s) cos(30))=3.849 s`

b) To calculate the height of the castle wall, we need to find the initial height

Since the ball hits the ground at the end of it's motion, we know that y=0

`y=y_(o)+v_(oy)*t+(1)/(2)gt^(2)`

`0=y_(o)+30sin(30)*(3.849)+(1)/(2)(-9.81)(3.849)^(2)`

Solving for
`y_(o)`

`y_(0)= -30sin(30)*(3.849)-(1)/(2)(-9.81)(3.849)^(2)=14.8575m`

c) At the highest point of its trajectory, we know that the cannonball stops ascending and began going down, that means:

`v_(y)=0`

`v_(x)=30cos(30)=15√(3)(m)/(s)`

d) Since the velocity in x is constant, the velocity just before the cannonball hits the ground is:

`v_(x)=30cos(30)=15√(3)`

To find y-velocity of the cannon ball's velocity we need to use the formula:

`v_(y) ^(2) =v_(oy) ^(2) +2g(y-y_(o))`

`v_(y)  =\sqrt{v_(oy) ^(2) +2g(y-y_(o))} =\sqrt{(30sin(30))^(2) +2(-9.8)(0-14.8575)}`

`v_(y)=22.7202 (m)/(s)`

e) It's attached