Kinematics Question: Afigther plane flies horizontally at a constant speed of 200m/s. It carries a cannon incl

Question

Kinematics Question: Afigther plane flies horizontally at a constant speed of 200m/s. It carries a cannon inclined at an angle of 30 degress below the horizontal.
(a) If it takes 10s for teh projectile which has a muzzle velocity of 500m/s (with respect to the plane) to reach a target on the ground, find the height at which the plane is flying.

I don't understand the significance of the phase 'with respect to the plane' like what will change without this phase and what will change with this phase? What is muzzle velocity?
I got an answer of 2990.5m, is it correct?

(b)How far ahead of the target must the pilot fire the cannon to get a direct hit?
I got an answer of 4330m but I didn't take into account the speed of the plane so is the answer correct?

(c)Find the horizontal and vertical compotents of the velocity of the projectile just before it hits the target.
I got horizontal: 433m/s and vertical: 348.1m/s. Is it correct?

argg this means that i'm wrong

Thanks for the detailed and clear explanations

Answer 1

muzzle velocity is the velocity of which the projectile leaves the barrel of the cannon. since the barrel is attached to the plane, the actual velocity of the projectile becomes the sum of muzzle and plane velocity.

A) if the cannon is inclined 30 degrees from the horizontal, then its vertical velocity is 500sin30. i term this 'u' the initial vertical velocity of the cannon. this cannon is still subject to gravity, so it will accelerate at 9.81m/s² vertically for 10s. use s=ut+0.5at² to find displacement (s) using known value for time (t) and initial velocity (u). i found the plane's height to be 2990.5m

B) we know that the cannon also has a horizontal component of velocity due to the muzzel. we also know that the aircraft is moving. so the actuall projectile velocity is 200+500cos30. now since this is a horizontal velocity, gravity does not change it, so we just mutilply this velocity by 10 (seconds) to get the horizontal distance it will travel while it is airbound. the distance is 6330m. you got the answer of 4330 because you did not incorporate the velocity of the aircraft.

C) horizontal component of velocity does not change because there is no acceleration, so this value is just 200+500cos30=633
vertical component is affected by gravity for 10seconds, so at landing, it is (500sin30)+(9.81)(10)=348.1

you pretty much got everything right except for the fact that you did not know that the cannon is travelling much faster than normal because the gun itself is moving with the aicraft.

Answer 2

The phrase "with respect to the plane" is for clarity. When the projectile comes out it is moving with a velocity of 500 m/s with respect to the plane. But the velocity with respect to someone on the ground would be the sum of the two velocity vectors. - 200m/s horizontally + 500 m/s at 30 degrees below horizon.

Keep trying and you should get the right result.

Answer 3

As the other posters stated, the muzzle velocity is the velocity of the bullet when it leaves the gun. W/R/T the plane means it is as measured as if the plane is stationary.

2990.5 m is correct for the altitude.

The bullet will fly a distance of 6330.17 m horizontally before striking the ground. Remember to add the velocity of the plane to the muzzle velocity horizontal component.

Your vertical velocity is correct at 348 m/s

The horizontal velocity is 633.0127 m/s

j

Answer 4

IN the front of it because no matter if there's no air resistance component to the bombs ahead momentum will be switched over to vertical momentum with techniques from the stress of gravity. The bomb will fly in an arc as a replace of horizontally.

Answer 5

velocity is relative to the present elements !..thus no standard formula !!!!!!!!!!!!!!!!!!!!!!