Question on Projectile Motion?

Question

My professor gave the class this question to solve in class. We were all unable to solve it and neither was he. (at least not on the spot)

A cannon is on top of a cliff 100m high. In the valley below is a radio tower. The tower is 40m from the cannon. The cannon shoots cannon balls at a speed of 100m/s. What angle must you aim the cannon to hit the tower?

Answer 1

Since you don't say how tall the tower is, presumably you want to strike it's base?

This is a straight-forward problem using the equations of motion. Break the velocity of a cannon ball into x and y components based on some angle(s). You know that during the time of flight, the x distance will be 40 m with a constant velocity v0x. Here are some statements you can make:

x = 40 = v0xt

y = -100 = 100 + v0yt + 1/2gt^2

v0x^2 + v0y^2 = 100 ^2

Looks like 3 independent equations with 3 unknowns... should be able to solve that.

The angle can be determined since you know that the tangent of the angle will be v0y/v0x

This would be so easy with pictures... ;-)

Aloha

Answer 2

You can solve this as follows:

1. Represent the x and y equations of motion as parametric in time:

x = v*cos(a)*t

y = v*sin(a)*t - (g*t^2)/2

2. Eliminate time by substituting the value of time from the first equation into the second:

t = x/(v*cos(a))

y = v*sin(a)*(x/(v*cos(a))) - (g*(x/v/cos(a))^2)/2

3. Simplify this a little and apply the identity:

1/(cos(a))^2 = tan(a)^2 + 1

y = x*tan(a) - (g*(x/v)^2*(tan(a)^2 + 1))/2

4. This is now quadratic in tan(a) so the two roots can be determined by the quadratic equation. They are:

tan(a) = 1/(g*x)(v^2 +- sqrt(v^4-2*g*y*v^2-a^2*x^2))

5. For your problem, x = 40 m and y = -100 m (the bottom of the tower). Your 2 solutions are 88.9 degrees and -67.1 degrees.

The first solution is almost straight up because your initial velocity is so high. The second solution is almost a straight shot at the tower base (that would be -68.2 degrees) because there is so little arc at that speed. If you reduce the initial speed, the two solutions finally meet each other (at v=8.69 m/s). Below that speed, no real solution exists.

Answer 3

The ball will move at 100m/s at w/e angle it is pointed.
It will also fall at 9.8m/ss. What you need then it how fast it will be traveling on the x axis to get your angle (right angled triangle, 9.8m/ss down(gravity), 100m/s on the h)
plug in the acceleration of 9.8m/ss and the distance of 100 m into a formula that gives you time. now you have time and distance, (40m and t) take it toss it into a formla that gives you V and toss it into a triangle, then, uh, I think it will work, I'm 90% sure the whole, pointing it up bit will just ballance out in the end but since it's also 100m up...hmm.
maybe it would be better to have the cannon pointing down? damn uh, i'm pretty sure there is just some trick in adding or subtracting 100 to your starting distance value(or any distance value you use) damn shame I JUST today deleted a diagram of a catapult I made that would have answered just this question, it's been a while since i took physics or any math, good luck!

Answer 4

More information is needed to answer this question. How tall is this tower? What part of tower do you want to hit?

I'm sure your professor realized this, but perhaps too late. Or you got the question wrong. Don't be too hard on him, because he was probably 5 steps ahead of you and did not realize the assumptions that would have to be made, until he tried to answer it in a way you would understand. It is a bad question, or perhaps you are the one who did not write it correctly. That is much more likely. Trust me, he knows more physics in his little finger than you will ever know in the rest of your life.

Come back and give the answer and this time ask a proper question.

Answer 5

that is a good one.

On the cliff express the velocity as Ux in the direction of X-axis and Uy in the direction of Y-axis. then
if Uo initial then
Ux=Uo cos(θ) and
Uy=Uo sin(θ)

Ux ideally is constant therefore for the whole motion the ball trevels 40m (horizontal speed) with a constant speed =>

Ux = S /t => t= S/Ux = 40/ Uo cos(θ)

Nopw for the vertical direction.:

the ball is traveling upwards to maximum height when its Uy becomes zero and then from this point it accelerates till it hits the tower trevelling 100m + H max
(Hmax measured from the canon)

so

t =t1 +t2
where t1 is from canon level to Hmax and then bck to canon level and
t2 is from canon level to tower (traveling for 100m) and
t is total time that equals to 40/Uocos(θ) since it fundamentaly the same since for the same tme the canon ball moves vertically and horizontally.

now all the kinetic energy of Uy in the cannon vecomes potential energy in Hmax.


therefore 1/2mUy^2 = mgHmax => Hmax known

.....
....
.....bla bla bla. algebra and mixing equations....

so you relat total times travel. vector of horizontal and vertical velocities, breal the motion in parts and eventually after some algebra you will get there.

Answer 6

Let the angle be x and u be the velocity then Horizontal velocity be u cos x= 100 cos x
Horizontal distance = 80 (given)
Distance = (velocity )(time)
80=(100cos x)t
or t =80/(100cos x)=0.8/cos x......................i
Vertical distance = 0
Vertical velocity = 100 sin x
by using the formula s=ut-1/2 gt^2
0=(100 sin x )t-0.5(9.8) t^2
0=t(100 sin x -4.9 t)
t=100 sinx/4.9................................................ii
Now elliminate t from i and ii
100 sinx/4.9=0.8/cos x
sin x cos x = (0.8)(4.9)/100=0.0392
2 sin x cos x = 0.0784
sin 2x = 0.0784
2x = sin^-1 (0.0784)
2x =4.496603602
x =2.25 degree

Answer 7

I'll give you a hint: There are *two* angles that will work ☺

And if your Prof. couldn't solve this one.....? What's his degree in, Physical Education or History?


Doug