Question

2) Projectile through a tube: A projectile is fired horizontally with speed v, from the top of a cliff of hcight h. It immediately enters a horizontal tube of length x. There is friction between the projectile and the tube such that the projectile loses velocity with contant acceleration -a. After leaving the tube, the projectile undergoes normal projectile motion to the ground a. What is the total horizontal distance l traveled by the projectile measured from the base of the cliff expressed in terms of x, h, b. What value of x vields the maximum value of 1?

Answer 1

The projectile with initial horizontal velocity
The projectile with initial horizontal velocity v_0 travels through the tube of length x with an acceleration - a to reach its final velocity at the end of the tube v = squareroot v^2_0 - 2ax (using v^2 - u^2 = 2as) With this horizontal velocity v the projectile falls from hill of height h. Horizontal distance travelled by projectile from the edge of tube to ground is (1 - x) in time t. horizontal distance = horizontal velocity x time taken to reach ground (1 - x) = ut ----(1) Vertical distance travelled by projectile from the edge of tube to ground in time t is h Using s = ut + 1/2 at^2, vertical distance h = 0 + 1/2 gt^2 (initial vertical velocity = zero) t = squareroot 2h/g (1 - x) = squareroot v_0^2 - 2ax squareroot 2h/g equation (1) can be written as That is l = x + squareroot v_0^2 - 2ax squareroot 2h/g b) For the maximum value of l, dl/dx = 0
travels through the tube of length with an acceleration to reach its final velocity at the end of the tube $v=\sqrt{v_0^2-2ax}$ ( using $v^2-u^2=2as$ )

With this horizontal velocity   the projectile falls from hill of height .

Horizontal distance travelled by projectile from the edge of tube to ground is in time .

horizontal distance = horizontal velocity x time taken to reach ground

$(l-x)=vt$ ----(1)

Vertical distance travelled by projectile from the edge of tube to ground in time is

Using $s=ut+\frac{1}{2}at^2$ , vertical distance   $h=0+\frac{1}{2}gt^2$ ( initial vertical velocity = zero)

$t=\sqrt{\frac{2h}{g}}$

equation (1) can be written as $(l-x)=\sqrt{v_0^2-2ax}\sqrt{\frac{2h}{g}}$

That is $\boxed{l=x+\sqrt{v_0^2-2ax}\sqrt{\frac{2h}{g}}}$

b) For the maximum value of , $\frac{dl}{dx}=0$

$\frac{d}{dx}\left ( x+\sqrt{v_0^2-2ax}\sqrt{\frac{2h}{g}} \right )=0$

$1+\frac{1}{2}(v_0^2-2ax)^{-1/2}(-2a)\sqrt{\frac{2h}{g}}=0$

$(v_0^2-2ax)^{-1/2}=\frac{1}{a}\sqrt{\frac{g}{2h}}$

$(v_0^2-2ax)=a^2\,\,{\frac{2h}{g}}$

$\boxed{x=\frac{v_0^2}{2a}-\,\,{\frac{ah}{g}}}$