Mechanics - Castle under siege (needs checking)

A castle is under siege. Its walls are 40m high, and it stands on level ground. The besiegers have a cannon at ground level 200m away and aim directly towards the castle and an angle of 35 degrees to the horizontal.

Using the standard equation for path, or working from first principles, find the speed of firing for a cannonball to just clear the top of the wall of the castle and find the time taken?

The defenders also have a cannon placed at the top of the walls, and can only fire this horizontally. Their aim is to either hit the opposing cannon, or to hit the members of the opposing army strung out behind for a distance of a further 200m. What range of values of speeds of firing is appropriate for the defenders’ cannonball?

PS: Air resistance is to be ignored in this question. g=acceleration du to gravity = 10

My answer: please could someone take 5 mins to check all calculations to see if I'm correct, and tell me if I have gone wrong somewhere. Thanks :-)

Right...

Solving forces horizontally:

constant speed of: u.cos alpha

so
x = u.cos alpha.t
200=u.cos 35. t (1)

Solving vertically:

a=-g
Initial velocity: u.sin alpha
y = u.t + 1/2 a.t^2
y = u. sin alpha . t - (1/2) g.t^2
40 = u. sin 35.t - (1/2) (10). t^2 (2)

From (1) we get u = 200 / cos 35.t

replacing in (2) we get:

40 = (200 / cos 35.t) sin 35.t - (1/2) g.t^2
40 = 200.tan 35 - (1/2) g.t^2
(1/2)g.t^2 = 200 tan 35 - 40
5t^2 = 200 tan 35 - 40
t^2 = 40 tan 35 - 8
t = 4.473 seconds

As
u = 200 / cos 35.t
u = 200 / cos 35 * 4.473
u = 54.584 ms-1

That checks out with my solution. (I did the same setup, but I solved equation 1 for t and substituted into equation 2. That way I got an equation for u directly. That means any errors in your Math would be different from mine. Since I got the same answer as you, we're cool. :) )

-Dan

Quote:

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Originally Posted by Natasha1

A castle is under siege. Its walls are 40m high, and it stands on level ground. The besiegers have a cannon at ground level 200m away....

The defenders also have a cannon placed at the top of the walls, and can only fire this horizontally. Their aim is to either hit the opposing cannon, or to hit the members of the opposing army strung out behind for a distance of a further 200m. What range of values of speeds of firing is appropriate for the defenders’ cannonball?

PS: Air resistance is to be ignored in this question. g=acceleration du to gravity = 10

u = 54.584 ms-1

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So, for this second part,...

The range horizontally is from 200m to 400m for the cannon ball to reach. So we just solve for the ball's speed to reach 200m, and the speed to reach 400m, and we're done.

Vertical:
No vertical component for the speed of the cannon ball, because it is fired horizontally. So,
y = 0*t +(1/2)g*t^2 --------y is taken as positive downwards, that's why the "+"
40 = (1/2)(10)(t^2)
t^2 = 40/5 = 8
t = sqrt(8) = 2sqrt(2) sec. ----the time that the ball will hit ground, at any initial horizontal speed.

For the 200m horizontal distance,
x = v*t
200 = v*2sqrt(2)
v = 200 /2sqrt(2) = 100/sqrt(2) = 50sqrt(2) m/sec ---***

For the 400m horizontal distance,
x = v*t
400 = v*2sqrt(2)
v = 400 /2sqrt(2) = 100sqrt(2) m/sec ---***

Therefore, the cannon balls have to be fired anywhere from 50sqrt(2) m/sec to 100sqrt(2) m/sec in order to hit the enemy.