Projectile motion

**Mtl** · April 9th 2008, 03:47 PM

I know Projectile motion is physics in nature but i am having problems with the mathematical mechanics and any help would be greatly appreciated,

A fighter pilot blows up a artillery dump. Diving at an angle of 53degrees below the horizon, the pilot releases a bomb at an altitude of 750m. The bomb hits the target 5.0s after being released.
What is the velocity of the bomber?

So i enter this into the equation d= vt - (1/2)gt^2 yet i get a negative number which to me doesnt make sense.

**Mathstud28** · April 9th 2008, 03:50 PM

Originally Posted by Mtl

I know Projectile motion is physics in nature but i am having problems with the mathematical mechanics and any help would be greatly appreciated,

A fighter pilot blows up a artillery dump. Diving at an angle of 53degrees below the horizon, the pilot releases a bomb at an altitude of 750m. The bomb hits the target 5.0s after being released.
What is the velocity of the bomber?

So i enter this into the equation d= vt - (1/2)gt^2 yet i get a negative number which to me doesnt make sense.

Did you consider that the equation for $V_x$ and $V_y$ are differnt...I hope I am not making a fool out of myself lol

**Mtl** · April 9th 2008, 03:58 PM

"Did you consider that the equation for

and

are differnt"

Yes. I should have defined all my variables a little better. The d = distance in the y direction (in metres), the v = initial velocity of the bomber (metres/second), t = time (seconds)

"...I hope I am not making a fool out of myself lol"

It was a valid question, thanks for your help.

**topher0805** · April 9th 2008, 04:01 PM

Originally Posted by Mtl

"Did you consider that the equation for

and

are differnt"

Yes. I should have defined all my variables a little better. The d = distance in the y direction (in metres), the v = initial velocity of the bomber (metres/second), t = time (seconds)

"...I hope I am not making a fool out of myself lol"

It was a valid question, thanks for your help.

I think that what he is trying to say is that you have set up you equation wrong. You can not just do it in one step. This is vector dynamics so velocity and acceleration have direction.

**topsquark** · April 9th 2008, 04:07 PM

Originally Posted by Mtl

I know Projectile motion is physics in nature but i am having problems with the mathematical mechanics and any help would be greatly appreciated,

A fighter pilot blows up a artillery dump. Diving at an angle of 53degrees below the horizon, the pilot releases a bomb at an altitude of 750m. The bomb hits the target 5.0s after being released.
What is the velocity of the bomber?

So i enter this into the equation d= vt - (1/2)gt^2 yet i get a negative number which to me doesnt make sense.

Define a coordinate system where the origin is on the ground directly below the plane when the bomb is dropped. I'm taking +x in the direction the plane is traveling and +y straight up.

We know for the bomb that
$x_0 = 0~m$ .... $x =$ ?
$v_{0x} = v_0~cos(53)$ .... $v_x =$ ?
$a_x = 0~m/s^2$

$y_0 = 750~m$ .... $y = 0~m$
$v_{0x} = -v_0~sin(53)$ .... $v_y =$ ?
$a_x = -9.8~m/s^2$

where $x, v_x, y, v_y$ are the positions, etc. at time t = 5.0 s, and $v_0$ is the speed the plane is moving at as it drops the bomb.

We have three main equations:
$x = x_0 + v_{0x}t$
$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$
$v_y = v_{0y} + a_yt$

See if that helps. (The basic idea is that we have the motion equations independent of each other in each coordinate direction.)

-Dan

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