# Homework Help: Cannon Projectiles

1. Jul 19, 2013

### postfan

1. The problem statement, all variables and given/known data
A cannon is shooting emergency packets to people stranded on the roof of a flooded building of height H = 106 meters relative to the cannon, the corner of which is located a distance D = 52 meters from the cannon. It is desired that the incoming packets are flying tangent to the roof as shown so that they land gently with as little impact as possible and slide along to a stop.

Find the initial speed v_0 and at what angle theta (in radians) the cannon should be aimed to achieve the above scenario.

2. Relevant equations

y=ax^2+bx+c

3. The attempt at a solution

I created a parabola in the form of y=ax^2+bx+c that goes through the points (0,0) and (52,106) with a=-4.9 (2nd derivative is acceleration) and then finding b using the first derivative getting b=v=256.838, but this isn't the right answer. Could anyone please tell me what I am doing wrong and how to fix it. Thanks!

2. Jul 19, 2013

### voko

Note that your equation of parabola is y = y(x). So the derivatives you are talking about are derivatives with respect to horizontal displacement x, not time, so 2a is not acceleration and b is not velocity.

3. Jul 19, 2013

### postfan

Could you please elaborate, I don't fully understand.

4. Jul 19, 2013

### voko

Which part do you not understand?

5. Jul 19, 2013

### szynkasz

Vertical and horizontal movement is described by equations:
$\begin{cases}x=v_0t\cos\theta\\y=v_0t\sin\theta-\frac{1}{2}gt^2\end{cases}$

Last edited: Jul 19, 2013
6. Jul 19, 2013

### postfan

Why isn't 2a the acceleration and why isn't b the velocity?

7. Jul 19, 2013

### voko

Acceleration and velocity are defined as derivatives with respect to time. x is not time.

8. Jul 19, 2013

### postfan

So then what is the acceleration and velocity?

9. Jul 19, 2013

### voko

Velocity is the time derivative of displacement. Displacement is the vector (x, y). Acceleration is the time derivative of velocity.

10. Jul 21, 2013

### PeterO

To meet your requirements, you want a projectile that, from a vertical point of view, reaches a maximum height of 102 m. On Earth that is true for only one vertical velocity.

In the time it takes to reach that maximum height, the horizontal distance of 52 m has to be covered.

You can use the usual vertical motion formulae to calculate the vertical speed needed.

I used to remember that for vertical projection, if the initial velocity is 10 m/s up, the max ht reached is a little over 5m.
20 m/s gives height just over 20m
30 m/s give 45+
40 m/s gives 80m+
50 m/s gives 125m+

From that I can guess that these projectiles must have a vertical velocity of ~ 45 m/s, and will take approx 4.5 seconds to reach maximum height.

Horizontally, that means they cover 52 m in 4.5 seconds - so at a velocity ~ 12 m/s

Adding those two components as vectors we get an initial speed around √(2025 + 144) ≈ 45.1

The tangent the angle above the horizontal will be approx. 45/12

If you use the usual equations you will find the actual vertical and horizontal speeds and should thus get the correct answer