Physics, 31.07.2019 18:20 antonjas001

Amedieval city has the shape of a square and is protected by walls with length 500 m and height 15 m. you are the commander of an attacking army and the closest you can get to the wall is 100 m. your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). at what range of angles should you tell your men to set the catapult

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The range of angles for attacking in medieval city follow the projectile motion. Thus angle should the army commander decide to set the catapult is $\theta\in(13^o,33.51^o)$ and $\theta\in(56.49^o,85.53^o)$ .

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

For the motion of horizontal and vertical direction we use following equation in projectile motion.

$x=x_0+u_0_xt\\y=y_0+u_0_yt-\dfrac{1}{2}gt^2$

Here, is the gravity and is time.

Given information-

The length of the wall is 500 m.

The height of the wall is 15 m.

The closest the army man can get to the wall is 100 m.

The Initial speed is 80 m/s.

For just hitting the wall,

The closest the army man can get to the wall is 100 m and initial speed is 80 m/s. Thus

$100=80\cos(\theta)\times t\\t= \dfrac{100}{80\cos\theta}$

Let the above equation as equation 1.

As the height of the wall is 15 m. Thus,

$15=80\sin (\theta)\times t-\dfrac{1}{2} (9.81)t^2\\15=80\sin (\theta)\times \dfrac{100}{80\cos\theta}-\dfrac{1}{2} (9.81)( \dfrac{100}{80\cos\theta})^2$

On solving the above equation we get the two values of the angle as,

$\theta=13^o, 85.53^o$

Now at this hits at the base of other side, the value of y is 0 and x is 600. Thus,

$600=80\cos(\theta)\times t\\t= \dfrac{600}{80\cos\theta}$

Let the above equation as equation 1.

As the height of the wall is 15 m. Thus,

$0=80\sin (\theta)\times t-\dfrac{1}{2} (9.81)t^2\\0=80\sin (\theta)\times \dfrac{600}{80\cos\theta}-\dfrac{1}{2} (9.81)( \dfrac{600}{80\cos\theta})^2$

On solving the above equation we get the two values of the angle as,

$\theta=33.51^o, 56.49^o$

Therefore the range of angles should the army commander decide to set the catapult is $\theta\in(13^o,33.51^o)$ and $\theta\in(56.49^o,85.53^o)$ .

Learn more about the projectile motion here;

link

Answer from: daltonrebekah3532

For hitting top of the front wall angle is 12.95° and 85.53°.

For hitting back of the wall angle is 35.71° and 55.22°.

Explanation:

The equation of projectile motion can be written as

x=u cosθt ,x=80 cosθ t

$y=usin\theta -\dfrac{1}{2}gt^2$

$y=80sin\theta t -\dfrac{1}{2}gt^2$

Now to find the angle for hitting top of the front wall so values of x and y will be x=100 m and y=15 m.

Now by putting the values

100=80 cosθ t

$t=\dfrac{1.25}{cos\theta}$

$15=80sin\theta t -\dfrac{1}{2}\times 9.81t^2$

$y=80\times1.25 \tan\theta -\dfrac{1}{2}\times 9.81(1.25 sec\theta)^2$

we know that $sec^2\theta=1+tan^2\theta$

So $15=100tan\theta-7.66(1+tan^2\theta)$

This is the quadratic equation .

$7.66tan^2\theta-100tan\theta+22.66=0$

So the values of angle will be 12.95° and 85.53°.

For hitting back of the wall

x=500 m and y=15 m.

600=80 cosθ t

$t=\dfrac{7.5}{cos\theta}$

$y=80\times7.5 \tan\theta -\dfrac{1}{2}\times 9.81(7.5 sec\theta)^2$

So our quadratic equation will be

$275.9tan^2\theta-600tan\theta+290.9=0$

So the values of angle will be 35.71° and 55.22°.

Answer from: Quest

distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

slope of a graph = change in value of y -axis / change in values of x -axis

comparing both the equations

change in value of y -axis = change in position

change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

Answer from: Quest

Letter a, because when there are unlike charges they will always attract to each other.

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