Kinematics involving an airplane

[davedave]

Homework Statement

A bomb is dropped from an airplane at an attitude of 14400 ft. The plane is moving at 600 miles per hour. How far will the bomb move horizontally after it is released from the plane?

Homework Equations

I use the formula involving the distance traveled by an object with no air resistance and the Pythagorean theorem.

(1) d= vt - 0.5gt^2
(2) a^2 + b^2 = c^2

The Attempt at a Solution

1 mile = 5280 ft
g = 32 ft/sec^2
the plane's speed = 600 miles per hours = 880 ft per sec

Assuming that the airplane has no vertical velocity, I let v = 0 in the first equation above.
Then, -14400 = -(0.5)(32)(t^2)
Solving for t gives t = 30 sec.

Next, the diagonal distance from the point at which the bomb is on the ground to the plane is
(880)(32)=28160 miles.

To find the horizontal distance traveled by the bomb, I use the Pythagorean formula

horizontal distance = square root of (28160^2 - 14400^2) = 24200 ft = 4.58 miles

My answer is wrong. The answer key says 5 miles. What did I do wrong?
Please explain it. Thank you very much

 

[grassstrip1]

Your work for the first part is correct the time is 29.9 seconds to reach the ground. For the second part what you did is incorrect the path is a parabola so the diagonal distance is not what you wrote there.

To find the horizontal displacement use d = v_xit + 1/2 a_x t²
You know the initial velocity in the x direction, and you know that the acceleration is zero since the only force acting on the body is the downward weight

 

[SteamKing]

The Attempt at a Solution

1 mile = 5280 ft
g = 32 ft/sec^2
the plane's speed = 600 miles per hours = 880 ft per sec

Assuming that the airplane has no vertical velocity, I let v = 0 in the first equation above.
Then, -14400 = -(0.5)(32)(t^2)
Solving for t gives t = 30 sec.

Next, the diagonal distance from the point at which the bomb is on the ground to the plane is
(880)(32)=28160 miles.

Always pay attention to the units you are using.

880 is in feet/second.
It's not clear if 32 represents g in ft/s² or if it is supposed to be the time of fall of 30 seconds.

In any event, the result does not come out in units of miles. 28160 miles is larger than the diameter of the earth!

 

[Mastermind01]

Next, the diagonal distance from the point at which the bomb is on the ground to the plane is
(880)(32)=28160 miles.

How did you get the diagonal distance by multiplying the horizontal speed with the gravitational attraction?

Anyway, the key thing here is that the horizontal distance traveled by the bomb only depends on the horizontal component of it's velocity. The horizontal distance is unaffected by the vertical component. Thus when the bomb leaves the plane it has a horizontal velocity of 880 ft/sec and the time for the journey as you found out was 30 sec. Therefore horizontal distance traveled is $880 \frac{ft}{s} * 30 s = 26400 ft = 5 miles$