Challenging Applied maths question projectiles

[lukesean]

An aircraft flies at a constant height H and constant velocity V. When the aircraft has flown directly over a gun on the ground a shot is fired from the gun which points at the aircraft at an angle of elevation Ѳ. If the initial velocity of the bullet is KVsecѲ [k >1], and Ѳ =tan inverse[ 1/V√gh/(k-1) ], show that the bullet hits the aircraft directly.

Please if anyone can answer me this question you can send it to me in an email <email addy removed by admin> if you make it on a paper please send me the pics. I appreciate it a lot.

If you have any questions regarding the probelem please feel free to contact me.

 

[MarkFL]

Hello and welcome to MHB. :D

We aren't a "problem solving" service...our main goal here is to help people solve problems by looking at what they have done and offering guidance to help them get unstuck, so that they are actively engaged in the process of coming to a solution. Response are posted in the thread started by the OP rather than sent by email, so I have removed your email address from public view.

So, if you can show us what you have tried, we can offer assistance aimed at helping you proceed.

 

[lukesean]

Hi Mark

I guess your from Florida.

Thanks for replying. I started the probelem by first applying the equation s=ut+1/2at2 equation vertically from the projected point. Next i found the time but from then onwards i got stuck with trying to process how to show that the bullet hits the aircraft directly

 

[MarkFL]

I would begin by obtaining the parametric equations of motion for the projectile (where we ignore the forces of drag):

Along the horizontal component of motion, there are no forces acting on the projectile, so we may state:

$$\frac{dv_x}{dt}=0$$ where $$v_{x_0}=v_0\cos(\theta)$$

Integrating with respect to $t$, we find:

$$v_x(t)=v_0\cos(\theta)$$

Integrating again, where the origin of our $xy$-axes is at the muzzle, we find:

$$x(t)=v_0\cos(\theta)t$$

Along the vertical component of motion the force of gravity is acting, in a downward direction, so we have:

$$\frac{dv_y}{dt}=-g$$ where $$v_{y_0}=v_0\sin(\theta)$$

Integrating with respect to $t$, we find:

$$v_y(t)=-gt+v_0\sin(\theta)$$

Integrating again, we get:

$$y(t)=-\frac{g}{2}t^2+v_0\sin(\theta)t$$

Can you state the equations of motion for the airplane, assuming the airplane is moving in a positive direction?