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A cannon is fired from the floor of a canyon at the vertical wall 3200 metres away. If the muzzle velocity is 200 m/s and the barrel makes an angle of 37 degrees with the horizontal, how high up the wall will the shell hit?

Question

A cannon is fired from the floor of a canyon at the vertical wall 3200 metres away. If the muzzle velocity is 200 m/s and the barrel makes an angle of 37 degrees with the horizontal, how high up the wall will the shell hit?

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Step 1

Here the figure shows a cannon projected from the origin with initial velocity u at an angle θ with the horizontal.

у
в - 37°
х
х3 3200 m
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у в - 37° х х3 3200 m

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Step 2

Given,

x =3200 m
u = 200 m/s
0 = 37°
9.8 m/s
also, g
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x =3200 m u = 200 m/s 0 = 37° 9.8 m/s also, g

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Step 3

Which has horizontal and vertical components as ux and...

a = 0
u =ucose
Fucos(370)
usin0,a, -
=-g
u,usin(37)
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Image Transcriptionclose

a = 0 u =ucose Fucos(370) usin0,a, - =-g u,usin(37)

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