Thread: projectile

a particle is projected inside a tunnel which is 8 m high with initial speed u. show that the max range inside the tunnel is 8[√ {(u^2 -16g)/g}]

so I got t=usina/g

sina=√(16g/u^2)

I then tried to sub this into distance in the x/horizontal direction but I could not rearrange it to get the right answer

$t = \dfrac{u \sin(a)}{g}$

is the time of maximum height, not maximum range.

As the starting and ending heights are the same the time of max range will be twice the time of max height.

So try subbing $t = \dfrac{2u\sin(a)}{g}$

into your formula for $x(t)$