whos smart in physics (Projectile Motion)

Is the plane directly above the car at the time the bomb is thrown?
Are they traveling in the same direction?

"At which below the horizontal should he throw the bomb so that it hits OSAMA??"

Which what? Angle? Velocity?

If the answers to the first two questions are yes, then he could throw the bomb at 37.5 m/s on the horizontal out the back of the plane. The bomb will always be over the car since the horizontal components of their respective velocities will be the same.

oopz
sorry...i forgot to writte some words...

its what angle from horrizontal should he throw the bomb. Secondaly u donn know from where they started

Rahul

The angle from the horizon doesn't matter.

The horizontal component of the bomb's velocity must match the car's. Add the horizontal component of the car's velocity to the opposite of the horizontal component of the plane's velocity. That will keep the bomb over the car until it hits it.

The vertical component of the bomb's velocity will only affect how long it takes for the bomb to hit the car.

First of all, we have to assume that Bush simply drops the bomb, since no other information was given. Therefore the bomb will have the horizontal speed of Bush's plane relative to the ground. The relative speed between Bush and bin Laden is 62.5 m/s - 25 m/s = 37.5 m/s. The amount of time that the bomb will take to fall 100 meters can be obtained from the standard formula:

s = (1/2) gt^2

where s = distance, g = acceleration due to gravity, and t = time.

First solve for time:

t^2 = 2s/g
t = SQRT(2s/g) = SQRT[(2)(100 m) / (9.8 m/s^2)]
= SQRT(20.408 s^2)
= 4.518 seconds

In 4.518 seconds, the 37.5 m/s relative speed between the plane and the car will carry the bomb (4.518 s)(37.5 m/s) = 169.4 meters horizontally towards the car. Therefore, Bush must release the bomb this distance behind bin Laden's car.

Assuming the plane is traveling in the same direction as the car.

Release would be conditional on the angle between the velocity vectors. As Brandon9000 determined the bomb will take 4.5185 seconds to fall (ignoring aerodynamic drag or lift).

So the solution really depends upon the angle between the magnitutude of the velocity vector of Osama's car and the airplane. So if you assume Osama's car to be traveling at phi degrees at 25m/s and the airplane to be traveling at 0 degrees and 62.5 m/s the solution is

D (m)=4.185(62.5-25Cos(phi))

Rap

Aside fom air friction, which is always ignored in Physics problems of this sort, the bomb will continue to travel with Bush horizontally as it falls, hence my solution. Any angle it might make with the ground is irrelevant.

righto

but osama's car wouldn't necessarily be traveling along the same vector as the airplane (and the bomb). So to aquire the target, the release point would have to take this angle between the two vectors into account.

Rap

If they weren't travelling in the same direction, it would be stated in the problem in any decently posed question. This appears to be just a very simple high school Physics problem without much subtlety.