Question

A bullet fired straight upwards experiences an acceleration due to air resistance of 1 m/s2 during its entire flight (this is an approximation; in reality it would depend on velocity.) If it leaves the barrel of the gun at 300m/s, with what speed does it hit the ground?

Answer 1

Answer:

$v_f=-270.86\ m/s$

Explanation:

Vertical Launch including Air Resistance

When an object is thrown in free air, the only acting force and acceleration are those of gravity. If air resistance is to be considered, then the effect is modeled as a constant value times the velocity or (like in this case) the acceleration.

The air resistance is modeled as an acceleration $a_r= 1\ m/s^2$ that must have the opposite direction of the movement, i.e. it goes down when the object is moving upwards and vice-versa.  

The net acceleration of the object when it's thrown up in the air is $-g-a_r$, both negative and adding because they brake the upwards movement of the object. When the object is returning to the ground, the acceleration is $-g+a_r$ because the resistance is opposing the movement.

Thus, the formula for the height of the bullet when it's moving upwards is

$\displaystyle y=vo.t-\frac{(g+a_r)t^2}{2}$

And the velocity is

$V_f=V_o-(a_r+g)t$

The bullet reaches its maximum height when Vf=0, thus

$V_o-(a_r+g)t_m=0$

Solving for tm

$\displaystyle t_m=\frac{V_o}{a_r+g}$

$\displaystyle t_m=\frac{300}{1+9.8}=27.78\ sec$

It reaches a maximum height of

$\displaystyle y_m=300\times 27.78-\frac{(9.8+1)\times 27.78^2}{2}$

$y_m=4,166.67\ m$

The bullet starts now its way down, with a net acceleration of $a_r-g$. The distance it travels at time t is given by

$\displaystyle y=-\frac{(a_r-g)t^2}{2}$

That distance is the maximum height already computed, thus

$\displaystyle -\frac{(a_r-g)t^2}{2}=4,166.67\ m$

Solving for t

$\displaystyle t=\sqrt{\frac{2\times 4,166.67}{9.8-1}}$

$t=30.78\ sec$

The final speed to reach the ground is

$v_f=(a_r-g)t$

$v_f=(-8.8)\times 30.78$

$\boxed{v_f=-270.86\ m/s}$

We can see part of the initial speed (and therefore kinetic energy) is lost due to air friction