1. ## Projectile motion (1)

An aeroplane flies horizontally at height , h in a straight line at a constant speed , u . When it is directly above a cannon on the ground , the cannon fires a shell which is supposed to hit the airplane . Ignore the air resistance , determine in terms of u,g and h the minimum muzzle speed of the shell , v.

Firstly , i resolve the velocity into 2 parts (ie $v_x$ and $v_y$)

$
v_x=u\cos \theta\Rightarrow v^2_x=u^2\cos^2 \theta
$

$
v^2_y=u^2+2as
$

$=u^2\sin^2 \theta-2gh$

$v=\sqrt{v^2_x+v^2_y}$

$
=\sqrt{u^2
-2gh}
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=\sqrt{u^2-2gh}
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The error here is in that sign , why should it be positive .

2. plane is travelling in a straight horizontal line with velocity u
let v be the initial velocity of cannon mounted at angle x
resolving velocity we get its
horizontal component=vcos(x)
vertical component=vsin(x)
since cannon ball hit the plane we have
vcos(x)=u (this will remain constant) equation 1
since we have to calculate the minimum cannon ball velocity so the final vertical component of velocity of cannon ball will be 0
therefor
(vsin(x))^2=2gh equation 2
squaring and adding equation 1 and 2 we have
(vcos(x))^2+(vsin(x))^2=u^2+2gh
therefor
v^2=u^2+2gh
and finally
v=(u^2+2gh)^(1/2)