Projectile Motion Homework: Plane Speed, Decoy Release & Displacement Equations

In summary, a plane with a speed of 180 mi/h and diving at an angle of 30.0 degrees released a radar decoy that struck the ground 2300 ft away. The plane was at a height of 1657.67m when the decoy was released and the decoy was in the air for 22.95 seconds. The acceleration of a sprinter running at 10 m/s when rounding a bend with a turn radius of 25m is 4 m/s2 and the acceleration vector points in an unknown direction.
  • #1
Sympathy
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Homework Statement


a certain airplane has a speed of 180 mi/h and is diving at an angle of 30.0 degrees below the horizontal when a radar decoy is released. The horizontal distance between the release point and the point where the decoy strikes the ground is 2300 ft. (a) How high was the plane when the decoy was released? (b) how long was the decoy in the air?


Homework Equations


Displacement equations


The Attempt at a Solution



I tried converting every distance unit to meters and got the velocity was 80.45 m/s
and the horizontal vel = 69.67 m/s
vertical vel = 40.225 m/s

but then I had to find part b before I could answer part a?

p.s. the answers I got were a) 1657.67m and b) 22.95 sec.

i don't think this is right
 
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  • #2
oh and 2nd problem

a) What is the acceleration of a sprinter running at 10 m/s when rounding a bend with a turn radius of 25m? b)In what direction does the acceleration vector point?

I got a), it's 4 m/s2. But what does b) asking?
 
  • #3
Your velocity components seems to be correct, but your dropping distance not.

Show us how you calculated a) with the substituted values. Hint: Use the horizontal speed to calculate the dropping time.
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air under the influence of gravity. It is a combination of horizontal and vertical motion, where the object follows a curved path known as a parabola.

2. How is plane speed related to projectile motion?

Plane speed is not directly related to projectile motion. However, the speed of the plane can affect the initial velocity of the projectile, which can impact the range and height of the projectile's path.

3. What is the significance of the decoy release in projectile motion?

The decoy release is a crucial moment in projectile motion as it marks the beginning of the projectile's flight. It is the point where the projectile's initial velocity and angle are determined, which ultimately determines the projectile's trajectory.

4. What are the displacement equations for projectile motion?

There are two displacement equations in projectile motion. The first is the horizontal displacement equation, which is given by x = v0 * t, where v0 is the initial velocity and t is the time. The second is the vertical displacement equation, which is given by y = v0 * t + 1/2 * a * t2, where a is the acceleration due to gravity.

5. How can projectile motion be applied in real-life situations?

Projectile motion has many real-life applications, such as in sports (e.g. throwing a ball, shooting a basketball), engineering (e.g. launching a rocket, designing roller coasters), and military operations (e.g. launching missiles, calculating artillery trajectories). Understanding projectile motion allows us to predict the path and landing point of a projectile, making it a useful tool in various fields.

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