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In a castle, overlooking a river, a cannon was once employed to fire at enemy ships.

One ship was hit by a cannonball at a horizontal distance of 150 m from the cannon.

The height of the cannon above the river was 67 m and the
cannonball was fired horizontally.

a)Show that the time taken for the cannonball to reach the water surface after being fired from the cannon was 3.7 s. Assume the air resistance was negligible.


b)Calculate the velocity at which the cannonball was fired.

c)Calculate the vertical component of velocity just before the cannonball hit the ship.

My answers:

For a) Resolving vertically u = 0

So s= ut + 1/2at2

s=1/2at2

67=1/2*9.81 * t2

t = 3.7

I get this.

But for b)

How do I know whether to calculate horizontally or vertically. If horizontally I can do it because a=0 so s=vt but how would I know.


Any help will be greatly appreciated.
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soup
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Note that in the question it says "the cannonball was fired horizontally". Thus there is no vertical component at t=0 only a horizontal component.
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(Original post by soup)
Note that in the question it says "the cannonball was fired horizontally". Thus there is no vertical component at t=0 only a horizontal component.
Thank you
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Yazanadi
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My man, X = Xu * t So Xu = X/t Xu = 67 / 3.7 Xu is the initial horizontal velocity X horizontal distance T is time
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maddy boy
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(Original post by Yazanadi)
My man, X = Xu * t So Xu = X/t Xu = 67 / 3.7 Xu is the initial horizontal velocity X horizontal distance T is time
u do realise that this thread is 6 years old
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