Projectile Motion airplane final speed

In summary, the problem involves an airplane dropping emergency supplies to survivors on an island. The airplane is flying at 116 km/hr and 914 m above the ocean. The final speed of the supplies upon landing on the island is determined using the equation Vfy^2 = Voy^2 +2ay, where Vx = 32.222 m/s, a = -9.8 m/s/s, Voy = 0 m/s, and y = -914 m. However, this only calculates the y-component of the velocity, so it must be combined with the x-component to determine the final speed.
  • #1
black_hole
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0

Homework Statement



An airplane flying 914 m above the ocean at 116 km/hr is supposed to drop a box of emergency supplies to the survivors of a shipwreck on an island. At what final speed will the supplies land on the island?


Homework Equations





The Attempt at a Solution



So Vx = 32.222 m/s, is Voy = 0 m/s?
a = -9.8 m/s/s and y = -914 m?

So I did Vfy^2 = Voy^2 +2ay and got -133.845 m/s, but that is not right...
 
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  • #2
black_hole said:
So I did Vfy^2 = Voy^2 +2ay and got -133.845 m/s, but that is not right...
That's just the y-component of the velocity. Combine it with the x-component to figure out the speed.
 
  • #3




Your initial attempt at a solution is on the right track, but there are a few errors in your calculations. Let's break down the problem and equations step by step to find the correct answer.

First, let's define our variables. Vx is the horizontal velocity of the airplane, which is given as 116 km/hr. We need to convert this to m/s, so Vx = 116 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 32.222 m/s. This part of your calculation is correct.

Next, we need to find the vertical velocity of the airplane, Voy. Since the airplane is flying at a constant altitude, we can assume that Voy = 0 m/s. This is because there is no change in altitude, so there is no vertical acceleration.

Now, we can use the equation Vfy^2 = Voy^2 + 2ay to find the final vertical velocity of the supplies, Vfy. We know that Vfy = 0 m/s, Voy = 0 m/s, and y = -914 m. Plugging these values in, we get 0 = 0 + 2(-9.8 m/s^2)(-914 m). Solving for Vfy, we get Vfy = 42.6 m/s.

Finally, we need to find the final speed of the supplies when they land on the island. This will be the resultant velocity, which is found using the Pythagorean theorem: Vfinal = √(Vx^2 + Vfy^2) = √(32.222 m/s)^2 + (42.6 m/s)^2) = 53.5 m/s.

Therefore, the final speed of the supplies when they land on the island is 53.5 m/s. This calculation takes into account the horizontal and vertical components of the velocity, and is the correct answer for this problem.
 

1. How is the final speed of an airplane in projectile motion calculated?

The final speed of an airplane in projectile motion is calculated using the equation Vf = Vi + at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time elapsed.

2. What factors affect the final speed of an airplane in projectile motion?

The final speed of an airplane in projectile motion is affected by the initial velocity, the angle of launch, the air resistance, and the force of gravity.

3. How does air resistance impact the final speed of an airplane in projectile motion?

Air resistance, also known as drag, acts in the opposite direction of the airplane's motion and decreases its speed. The greater the air resistance, the slower the final speed will be.

4. Can the final speed of an airplane in projectile motion be greater than its initial speed?

Yes, it is possible for the final speed of an airplane in projectile motion to be greater than its initial speed. This can happen if the angle of launch is greater than 45 degrees and there is minimal air resistance.

5. How does the final speed of an airplane in projectile motion change with increasing time?

The final speed of an airplane in projectile motion will continue to increase as time increases, as long as there is no significant air resistance. However, the rate at which the speed increases will decrease as the airplane approaches its maximum height, and then decrease again as it descends due to the force of gravity.

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