Question

Help asap please?

During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +35.0 m/s at an angle of 60.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.

Answer 1

I remember this question

http://openstudy.com/study#/updates/52cb2c88e4b047f54eb2fe2a

Which part are you stuck on?

Answer 2

basically the whole thing, i just need it to be explained step by step. @broken_symmetry so i can get it.

Answer 3

Parts of the question deal with how high the stone will go and how long it will stay in the air.
For these parts you only need to think about the vertical component of velocity.

The other part deals with horizontal distance from the castle, so for this part you only need to deal with the horizontal component of velocity.

Can you work out what the values of the horizontal and vertical components of velocity are?

It's the same mathematics as working out the sides of a right-angle triangle.

Answer 4

Once you have the initial vertical velocity, you can work out how long it stays in the air for.

The equation for displacement (s) given an initial velocity (u) and acceleration (a) as a function of time (t) is:
\[s = ut + \frac12at^2\]
u is the vertical component of velocity.
The acceleration is just the acceleration due to gravity. Make a note of the minus sign because it is acting downwards and we are calculating upwards displacement.
\[a = -g = -9.8\text{ m s}^{-2}\]
Lastly we know the displacement when the stone will come back to the ground again. It is zero. In vertical terms, the stone is back where it started.

Now we have a quadratic equation in time:
\[ 0 = ut -\frac12gt^2\]

Do you know how to solve quadratic equations?

Answer 5

Once you have solved this equation and found time-of-flight, you can find the maximum range. This is purely the time-of-flight multiplied by the horizontal component of velocity.

Answer 6

To get how high the stones go, you can use the equation for displacement and halve the time of flight, because that's the point in time when the stone is at the top of the parabola.

\[s = u\frac t2+\frac18at^2\]