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:

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

Integrating with respect to $t$, we find:

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

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

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

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

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

Integrating with respect to $t$, we find:

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

Integrating again, we get:

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

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

 

[CellCoree]

Dear reader,

Thank you for your inquiry regarding the challenging applied math question about projectiles. I am happy to provide a response to help clarify the problem and its solution.

The problem states that an aircraft is flying at a constant height H and velocity V. A shot is fired from a gun on the ground at an angle of elevation Ѳ, with an initial velocity of KVsecѲ [k>1]. The goal is to show that the bullet will hit the aircraft directly.

To solve this problem, we need to use the equations of motion for projectile motion. These equations describe the motion of an object under the influence of gravity, which is the case for a bullet fired from a gun.

First, let's define the variables used in the problem:

H = height of the aircraft
V = velocity of the aircraft
Ѳ = angle of elevation of the shot
K = constant
g = acceleration due to gravity

Now, let's break down the problem into smaller steps:

1. Find the equation for the horizontal distance traveled by the bullet.

The horizontal distance traveled by the bullet can be calculated using the equation:

d = Vt

where d is the horizontal distance, V is the initial velocity of the bullet, and t is the time it takes for the bullet to reach the aircraft.

2. Find the equation for the vertical distance traveled by the bullet.

The vertical distance traveled by the bullet can be calculated using the equation:

h = Vt + ½gt^2

where h is the vertical distance, V is the initial velocity of the bullet, and t is the time it takes for the bullet to reach the aircraft.

3. Substitute the values of V and t into the equations.

Since the aircraft is flying at a constant velocity V, the time it takes for the bullet to reach the aircraft is the same as the time it takes for the aircraft to fly over the gun. Therefore, we can substitute the value of t in both equations with H/V.

Substituting these values, we get:

d = V(H/V) = H

h = VH/V + ½g(H/V)^2 = H + ½g(H/V)^2

4. Use the given equation for Ѳ to simplify the equation for h.

The given equation for Ѳ is:

Ѳ = tan inverse[ 1/V√gh/(k-1) ]

Simplifying this equation, we get:

Ѳ =