# Thread: Projectile Motion and Impact

1. ## Projectile Motion and Impact

1. The problem statement, all variables and given/known data
A gun on the shore (at sea level) fires a shot at a ship which is heading directly toward the gun at a speed of 40 km/h. At the instant of firing, the distance to the ship is 15,000 m. The muzzle velocity of the shot is 700 m/s. Pretend that there is no air resistance.
(a) What is the required elevation angle for the gun? Assume g = 9.80 m/s^2.
(b) What is the time interval between firing and impact?

2. Relevant equations

Uhmm, I'm guessing. $t_{flight} = \frac{2v_{0} sin\alpha }{g}$

Maybe $x_{max} = \frac{v^{2}_{0}sin2\alpha}{g}$ will also be applicable.

$v_{x} = v_{0x} = v_{0}cos\alpha$
$v_{z} = v_{0z} - gt = v_{0} sin \alpha - gt$
$x = v_{0x}t$
$z = v_{0z}t - \frac{1}{2} gt^2$

3. The attempt at a solution
I'm not sure what to do with this problem. Any way I set it up, I end up with more variables than I can solve for.

I would appreciate a hint to throw me in the right direction. Thanks in advance.

2. Originally Posted by Jacobpm64
1. The problem statement, all variables and given/known data
A gun on the shore (at sea level) fires a shot at a ship which is heading directly toward the gun at a speed of 40 km/h. At the instant of firing, the distance to the ship is 15,000 m. The muzzle velocity of the shot is 700 m/s. Pretend that there is no air resistance.
(a) What is the required elevation angle for the gun? Assume g = 9.80 m/s^2.
(b) What is the time interval between firing and impact?
First I'm going to convert the speed of the ship to m/s:
$40~km/h \approx 11.111~m/s$.

I'm going to put an origin at the point of firing and use a +x direction in the direction of the oncoming ship and +y directly upward.

The key to doing this is you have to aim the projectile at where the boat is going to be when the projectile gets to the water level. So the range of the projectile is going to have to be:
$x = 15000 - 11.111t$ (in m).

From the projectile part of the problem we know that
$x = x_0 + v_{0x}t = v_0 \cdot cos(\theta) t$

Combining these two gives us:
$15000 - 11.111t = v_0 \cdot cos(\theta) t$

Again, from the projectile part of the problem we know that the final height of the projectile is the same as the initial, so
$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

$0 = v_0 \cdot sin(\theta)t - 4.9t^2$

So we have two equations in two unknowns to solve:
$15000 - 11.111t = v_0 \cdot cos(\theta) t$
and
$0 = v_0 \cdot sin(\theta)t - 4.9t^2$

There are two typical ways to solve this equation:
1) Solve the x equation for cosine and the y equation for sine, then use $sin^2(\theta) + cos^2(\theta) = 1$.
2)Solve the x equation for t and substitute into the y equation. Then solve the trigonometric equation.

I'm going to use 1):
$cos(\theta) = \frac{15000 - 11.111t}{v_0t}$
and
$sin(\theta) = \frac{4.9t^2}{v_0t} = \frac{4.9t}{v_0}$

So
$sin^2(\theta) + cos^2(\theta) = 1$

$\left ( \frac{4.9t}{v_0} \right )^2 + \left (
\frac{15000 - 11.111t}{v_0t} \right )^2 = 1$

$\frac{24.01t^2}{v_0^2} + \frac{(15000 - 11.111t)^2}{v_0^2t^2} = 1$ <-- Multiply both sides by $v_0^2t^2$:

$t^4 + (15000 - 11.111t)^2 = v_0^2t^2$

$t^4 + (123.454t^2 - 333330t + 225000000) = v_0^2t^2$ <-- Now I'm going to plug in $v_0 = 700~m/s$ and simplify:

$t^4 - 4898777t^2 - 333330t + 225000000 = 0$

We can solve this numerically, and I get that:
$t = 699.924~s, 21.1032~s$
(These are the only two positive solutions, the negative solutions being unphysical.)

So now we can find the projection angles:
$sin(\theta) = \frac{4.9 \cdot 699.924}{700} \approx 4.89947$
which is impossible.

$sin(\theta) = \frac{4.9 \cdot 21.1032}{700} \implies \theta \approx 8.495^o$

-Dan

3. Originally Posted by Jacobpm64
1. The problem statement, all variables and given/known data
A gun on the shore (at sea level) fires a shot at a ship which is heading directly toward the gun at a speed of 40 km/h. At the instant of firing, the distance to the ship is 15,000 m. The muzzle velocity of the shot is 700 m/s. Pretend that there is no air resistance.
(a) What is the required elevation angle for the gun? Assume g = 9.80 m/s^2.
(b) What is the time interval between firing and impact?

I was driving home a while back when this your question flashed to my mind.

Let me show anyway another way.

Since there is no air/wind resistance, the projectile will travel horizontally
(Vo)cos(alpha) *T
And the ship will travel (40kph)*T
for the two to meet, at impact.
So,
(Vo)cos(alpha) *T +(40kph)*T = 15,000 ---------(i)

Since g = 9.8 m/sec/sec, Vo = 700m/sec, thenwe need to convert the 40kph into m/sec;

(40km/hr)(1000m/1km)(1hr/3600sec) = 40(1000)/3600 = 11.111 m/sec

So,
700cos(alpha)*T +(11.111)T = 15000 ----------(ii)

What is T?

T is the time in seconds when the projectile hits the ship. It is the time when the projectile returns to the sea level.
The flight of the projectile is a parabola. At the vertex, the net vertical velocity of the projectile is zero. By property of parabola, the time, T, the projectile returns to the level of firing---the sea level---is twice the the time the projectile spent in reaching the vertex.

At the vertex,
(Vo)sin(alpha) -g(t) = 0 <----analogous to Vfinal = Vo +at.
700sin(alpha) -9.8t = 0
t = (700/9.8)sin(alpha) = 71.4286sin(alpha) sec.

T = 2t = 142.857sin(alpha) sec.

Substituting that into (ii),
700cos(alpha)*T +(11.111)T = 15000 ----------(ii)
700cos(alpha)*(142.857sin(alpha)) +(11.111)(142.857sin(alpha)) = 15000
100,000cos(alpha)sin(alpha) +1587.284sin(alpha) = 15,000
50,000sin(2 alpha) +1587.284sin(alpha) = 15,000 ---------------(iii)

There, a single equation in alpha to solve your problem.

I solve for alpha by iteration. Newton's Method is popular but you may not be into Calculus yet, so let's use my old and reliable iteration method.
Just test values of alpha on (iii).
It's fast if you are used to it.

After some trials, I found alpha = 8.59 degrees. -------------answer.

50,000sin(2*8.59deg) +1587.284sin(8.59deg) =? 15,000
15,005.8 =? 15,000
Almost, so, Ok.

-----------------------------
Time interval between firing and impact?

That is T.

T = 142.857sin(alpha)
T = 142.857sin(8.59deg) = 21.34 seconds ----------answer.

4. I should add a comment to ticbol's solution: he did it well, but there is a subtlety that can easily get missed. What ticbol got as an answer for the angle is a reference angle. In other words he found a solution $\alpha$, but another possible solution is $180^o - \alpha$. In this case the second "solution" doesn't work. However if the target were stationary it would be a valid (if somewhat bizarre) solution.

-Dan