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Welcome to the article During a fireworks display a shell is shot into the air with an initial speed of 75 709 m s at an angle of 6. On this page, you will learn the essential and logical steps to better understand the topic being discussed. We hope the information provided helps you gain valuable insights and is easy to follow. Let’s begin the discussion!

During a fireworks display, a shell is shot into the air with an initial speed of 75.709 m/s at an angle of 6.789 degrees above the horizontal. The fuse is timed to explode just as it reaches its highest point above the ground.

What is the horizontal displacement when it explodes? Use [tex]g = 9.8 \, \text{m/s}^2[/tex].

Answer :

Final answer:

To find the horizontal displacement of a firework shell at its peak, calculate the horizontal velocity component, determine the time to reach the peak by using the vertical velocity component and gravity, and then use the time and horizontal velocity to find the displacement.

Explanation:

The question involves calculating the horizontal displacement of a fireworks shell which is shot into the air at an angle and explodes at its apex. To solve this, we need to use the kinematic equations of projectile motion. The formula for horizontal displacement is \x = v_{x} t\, where \v_{x}\ is the initial horizontal velocity and \t\ is the time of flight to the apex. Given that the shell's initial speed is \75.709 m/s\ and the angle of projection is \6.789 degrees\, we first need to calculate the initial horizontal component of the velocity, \v_{x} = v \cos(\theta)\. We also find the time \t\ that the projectile takes to reach its highest point using the vertical component of the motion and the acceleration due to gravity g. However, since the question provides data from a different example, we could use that example for an understanding, but we will calculate the horizontal displacement using the given initial speed and angle for the problem at hand.

To solve for the horizontal displacement when the shell explodes, use the steps below:

  1. Calculate the horizontal component of the initial velocity: \v_{x} = 75.709 \cos(6.789^\circ)\.
  2. Calculate the vertical component of the initial velocity: \v_{y} = 75.709 \sin(6.789^\circ)\.
  3. Calculate the time \t\ to reach the highest point (where vertical velocity is zero) using \v_{y} = v_{oy} - gt\, where \v_{y} = 0\, and \g = 9.8 m/s^2\.
  4. Using the time \t\, calculate the horizontal displacement \x\ using \x = v_{x} t\.

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Rewritten by : Brahmana