Find the angle the bomb is dropped?

Question

A dive bomber has a velocity of 280 m/s at angle 'theta' below the horizontal. When then altitude is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of displacement from the release of the bomb to the target is 2250m, find theta.

I usually do not ask from homework help, but I have been working on this one for a while and I am still not sure how to approach it. Thanks everyone.

I am really thinking the problem is more complicated this. This is a college level problem, and the question is marked as difficult. The motion of the bomb is going to become parabolic very quickly, and I think the component velocities will need to be found. I know sometimes unneeded information is given, but I really do not think the answer is as easy as inverse sine (2150/3250)

Answer 1

This one looks like fun, so it will be the last one I do before going to sleep. The only question is what "magnitude of displacement" means ... I trust it means a straight line rather than length of the parabolic arc.

You have a right triangle ABC with the release point A at (0,2150); the right angle B at (0,0), and the target at point C on the ground. We know AC = 2250, so BC = sqrt(2250^2 - 2150^2) = sqrt(440000) = 200 sqrt(11). That's the distance from the release point to the target.

Break the problem into horizontal and vertical components. We'll set up the horizontal part first. Let vh be the horizontal component of initial velocity. (And horizontal velocity is constant.) Then

200 sqrt(11) = vh t = (vo cos theta) t = 280t cos theta (Eq 1)

where t is time and vo = 280 m/s is initial velocity.

Now do the vertical component using s = 1/2 gt^2 + vv t + so where s = 0 is the ground where it hit; g = -9.8 m/s/s; vv is the vertical component of initial velocity (a negative value); and so = 2150 is the initial vertical displacement.

0 = 1/2 (-9.8) t^2 - (280 sin theta)t + 2150
4.9 t^2 + 280t sin theta - 2150 = 0 (Eq 2)

From Eq 1,

t = 200 sqrt(11) / (280 cos theta) = 5 sqrt(11) / (7 cos theta)
t^2 = 275/(49 cos^2 theta)

Plug this into Eq 2:

4.9 [275/(49 cos^2 theta)] + 280 sin theta [5 sqrt(11) / (7 cos theta)] - 2150 = 0

This looks like a mess, but hopefully we can simplify:

27.5 / cos^2 theta + 200 sqrt(11) tan theta - 2150 = 0

Multiply through by cos^2:

27.5 + 200 sqrt(11) sin theta cos theta - 2150 cos^2 theta = 0

This still looks like a mess. There are a couple of double-angle trig identities that may help:

1) sin 2x = 2 sin x cos x
sin x cos x = 1/2 sin 2x

2) cos 2x = 2 cos^2 x - 1
cos^2 x = 1/2 (1 + cos 2x)

Let's make those substitutions and see what we get:

27.5 + 200 sqrt(11)[1/2 sin (2 theta)] - 2150 {1/2 [1 + cos (2 theta)]} = 0

27.5 + 100 sqrt(11) sin (2 theta) - 1075 - 1075 cos (2 theta) = 0

100 sqrt(11) sin (2 theta) - 1075 cos (2 theta) = 1047.5

Let's try for a numerical solution. Dividing out the 100 sqrt(11), we have

sin (2 theta) - 3.24125 cos (2 theta) = 3.15833

This is like sin x - 3.24125 cos x = 3.15833

I made a table of values for this, and found x = 141.46 degrees works well. sin 141.46 - 3.24125 cos 141.46 = 3.15828, which is close enough.

Since that equals 2 theta, the angle theta is 70.73 degrees. Round that off to 70.7 degrees for your answer.

That's a very steep angle, so it's clearly a dive bomber.

Now let's do a check. Since we know theta,

t = (5 sqrt 11) / 7 cos theta = 7.18 seconds

The horizontal and vertical velocity components are:

vh = vo cos theta = 280 cos 70.73 = 92.41 m/s
vv = vo sin theta = 280 sin 70.73 = 264.31 m/s

The horizontal distance it will travel is 7.18*92.41 = 663.5 m.

By the Pythagorean Theorem, the total displacement is

sqrt(663.5^2 + 2150^2) = sqrt(5062732) = 2250.05 m

which rounds off to 2250 m.

That proves the result.

Good problem, especially the trig substitutions and the numerical solution.

Answer 2

Well, you know the height of release (2150 m) and the horizontal distance it traveled (2250 m). You also know the speed it was released at (280 m/s). You'll need to use the x component and the y component of the velocity, just using cos and sin theta.

Try solveing for time in terms of theta using the x = x + vt equation, then plug that expression into the y equation. The math might get a little hairy (you'll probably need a trig substitution to solve it) but it's doable.

Hope this helps!

Answer 3

Draw a picture.

Include the ground (at the bottom), the horizontal (parallel to the ground), The path of the divebomber (extended all the way to the ground to make it easier).
Since the lines are parallel, the angle "theta" (formed by the horizontal and the path of the bomber) will be equal to the alternate interior angle formed by the ground and the path of the bomber. Form this bottom portion off into a triangle, with a height of 2250 m (2.25 km).

From here, you should be able to use some basic trignometric functions to figure out the angle theta, by using the inverse of sine, cosine, or whatever you decide on.

Answer 4

relies upon on the suggestions-set of the airplane. If the airplane dropping it rather is flying point, the bomb will drop point. considering which you're speaking approximately "dumb" iron bombs, then they does not have fins or different attachments that alter the steadiness or aerodynamics of the bomb too lots. If the airplane is diving, then needless to say the nostril of the bomb will factor with the airplane. the direction it takes is a ballistic one. A Mk. eighty two bomb does not have fins to "flow". It drops like a rock. That explodes. the hearth administration laptop computes the ballistic direction the bomb will take consistent with airplane speed and altitude and ballistic traits of the bomb. otherwise, if it glided, it may be stressful to come to a decision with the aid of fact it rather is unguided. There are bombs that flow to their objective. like the JSOW (Joinst Stand Off Weapon) that could strengthen the stand off variety of the airplane to keep away from SAMs and AAA.

Answer 5

Velocity 280