# Homework Help: Need help with Projectile Motion Questions

1. Jun 10, 2012

### lilmase

I need some help explaining these concepts

1. Suppose a tank is able to point its barrel up at an angle of 90o. If the tank is in motion and fires a shell straight up, where will the shell land if the tank maintains a constant motion. (don't consider any air resistance) a) in front of the tank b) behind the tank c) on top of the tank

2. Suppose an airplane drops a package out of its cargo hold while it is travelling through the sky with a constant horizontal speed. Where will the package land once it hits the ground? a) directly below the plane b) below the plane and ahead of it c) below the plane and behind it

My explanations
1. C. Because the shell does not accelerate in the x-direction and... (i'm lost here)

2. A. (Not sure why, intuitively I can see why it happens but cannot explain why)

Thanks.

2. Jun 10, 2012

### NegativeDept

The shell has an initial velocity, and that initial velocity has $x$ and $y$ components. They are:
$v_x =$ the horizontal velocity of the tank
$v_y =$ whatever the muzzle velocity of the gun is
Neglecting air resistance, the equations of motion for the shell can be found by using $\mathbf{F} = m \mathbf{a}$. The equations are: $\frac{d}{dt}v_x = 0$ and $\frac{d}{dt}v_y = - g$

Or, in words: $v_x$ stays constant because the $x$-component of force on the shell is zero. But the $y$-component of force is $-mg$, so the shell has a $y$-acceleration of $-g$. (I'm assuming you've defined the positive $y$-direction to be "up.") Gravity makes $v_y$ drop to zero, then become negative, then become really negative. But throughout its flight, the shell still has the same $v_x$ as the tank. So if the tank doesn't turn or change speed, it will be hit by its own shell.

If we neglect air resistance, it's a similar problem: the $v_x$ component of the package is constant until it hits the ground. (The $x$-component of gravity is zero, and there are no other forces, so its $x$-acceleration is zero.) If the $v_x$ component of the airplane also stays constant, the package will land directly under the airplane.

In a real-life air drop, the package will experience lots of drag force in the $-x$ direction. So will the airplane, but it uses its engines to compensate and keep $v_x$ roughly constant. In that case, the package would land behind the airplane.

3. Jun 10, 2012

### azizlwl

It relative speed we are dealing with.
If no external forces acted on objects then the velocity remain.

1. Ans C. The horizontal speed of the shell and tank are the same. Thus for any instant their relative position is constant that shell is always above the tank.

2. Ans A. As above their relative horizontal velocity is zero. The package lands directly above the plane since time taken and speed are the same with no external forces. After the object landed external force(friction) determine the position.

4. Jun 10, 2012

### lilmase

Thanks guys, I get it now, the velocity of the shell from the tank will keep moving with the same speed as the tank, and I'm guessing that's why it'll land right on top of it.