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In a Fourth of July celebration, a firework is launched from ground level with an initial velocity of 25.0 \(\mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) from the vertical. At its maximum height it explodes in a starburst into many fragments, two of which travel forward initially at 20.0 \(\mathrm{m} / \mathrm{s}\) at \(\pm 53.0^{\circ}\) with respect to the horizontal, both quantities measured relative to the original firework just before it exploded. With what angles with respect to the horizontal do the two fragments initially move right after the explosion, as measured by a spectator standing on the ground?

Short Answer

Expert verified
Fragments move initially at angles \( \theta_1 = 71.6^{\circ} \) and \( \theta_2 = 43.6^{\circ} \) from the horizontal.

Step by step solution

01

Break Down the Initial Velocity

The firework is launched at an angle of \( 30^{\circ} \) from the vertical. To work with horizontal angles, let's convert it to the angle from the horizontal: \( 90^{\circ} - 30^{\circ} = 60^{\circ} \). The firework is thus launched at \( 60^{\circ} \) from the horizontal with a velocity of \( 25.0 \, \text{m/s} \).
02

Determine Horizontal Velocity Component

The horizontal velocity component (v_x) of the firework just before the explosion can be calculated as: \[ v_x = 25.0 \, \text{m/s} \times \cos(60^{\circ}) = 12.5 \, \text{m/s} \].
03

Calculate Velocity Components of Fragments

Each fragment travels at \( 20.0 \, \text{m/s} \) at angles of \( \pm 53^{\circ} \) relative to the firework's velocity just before the explosion. The new horizontal components for the fragments are: \[ v_{x1} = 12.5 \, \text{m/s} + 20.0 \, \text{m/s} \times \cos(53^{\circ}) \] and \[ v_{x2} = 12.5 \, \text{m/s} - 20.0 \, \text{m/s} \times \cos(53^{\circ}) \].
04

Calculate Vertical Velocity Components for Fragments

The vertical components are: \[ v_{y1} = 20.0 \, \text{m/s} \times \sin(53^{\circ}) \] and \[ v_{y2} = 20.0 \, \text{m/s} \times -\sin(53^{\circ}) \].
05

Calculate Resultant Velocities

Use the Pythagorean theorem to find the resultant velocities of each fragment: \[ v_{1} = \sqrt{v_{x1}^2 + v_{y1}^2} \] and \[ v_{2} = \sqrt{v_{x2}^2 + v_{y2}^2} \].
06

Determine Angles Relative to the Horizontal

The angles \( \theta_1 \) and \( \theta_2 \) with respect to the horizontal are found using the tangent function: \[ \theta_1 = \arctan\left(\frac{v_{y1}}{v_{x1}}\right) \] and \[ \theta_2 = \arctan\left(\frac{v_{y2}}{v_{x2}}\right) \].
07

Solve for the Angles

Use the computed components from Steps 3 and 4 to find \( \theta_1 \) and \( \theta_2 \), yielding the angles with respect to the horizontal as observed from the ground.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Firework Trajectory
Understanding the trajectory of a firework is crucial when discussing projectile motion. A trajectory is the path that a projectile follows through space, and in the case of a firework, it begins at the launch site and continues to its point of explosion. This path is affected by gravity, launch angle, and initial speed.

When a firework is launched, its trajectory can be broken into two components: vertical and horizontal. These components are influenced by the angle of projection and initial velocity. For example, in the given exercise, the firework is launched at an angle of 60 degrees from the horizontal with an initial velocity of 25.0 m/s.
  • Parabolic Path: Fireworks typically follow a parabolic trajectory due to the influence of gravity, creating a symmetrical arch before reaching their peak height, where events like explosions usually happen.
  • Impact of Velocity: The initial velocity affects both range (the horizontal distance traveled) and the height (the maximum height reached) of the projectile.
By understanding the trajectory, we can predict the explosion's impact, helping us determine where fragments may land.
Velocity Components
In projectile motion, velocity is split into horizontal and vertical components. These components help determine the firework’s path and eventual points of impact.

For the firework problem, breaking down the velocity is essential:
  • Horizontal Component: It remains constant if we disregard air resistance. Initially, you can calculate it by using the cosine of the launch angle. For the given problem, the horizontal velocity component just before the explosion is calculated as 12.5 m/s.
  • Vertical Component: This component changes continuously due to gravitational acceleration. Initially derived using the sine of the launch angle, it influences how high the firework will climb before it descends back to the ground.
These components assist in predicting the exact path and help compute the effect of gravity on the firework.
Angles of Projection
Angles of projection significantly influence the motion characteristics of a projectile. In our context, this refers to the angle at which the firework is launched and how it affects the subsequent motion and impact.
  • Angle from Horizontal: When a firework is launched, it's crucial to convert the given angle from vertical to horizontal, as most trajectory calculations are easier with respect to the horizontal plane. For our problem, the conversion results in a 60-degree launch angle.
  • Resulting Effect on Trajectory: The exact angle affects both range and height. A greater angle from the horizontal typically results in a higher peak but a shorter range, while a smaller angle results in a longer distance but a lower peak.
In fireworks displays, selecting the correct angle ensures safety and maximizes the visual effect by dictating where and when the burst occurs.
Horizontal and Vertical Motion
Projectile motion can be broken into two independent parts: horizontal and vertical motions. These govern how a projectile, like a firework, travels through space, combining to form the resultant path.
  • Horizontal Motion: Constant as it is unaffected by gravity in ideal conditions. In the given scenario, the horizontal motion of fragments after explosion is modified by their initial velocities (e.g., 20 m/s at ±53° relative to the horizontal).
  • Vertical Motion: Influenced greatly by gravity, this involves upward deceleration until reaching the maximum height and then accelerating downwards. The vertical motion directly affects the visual height of the fireworks and their explosion timing.
By analyzing these motions separately, complex paths become easier to understand and calculate, allowing fireworks artists to predict where explosions will occur in the sky.

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Most popular questions from this chapter

Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at 3.05 \(\mathrm{m} / \mathrm{s} .\) Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is 38.0 \(\mathrm{m}\) above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally 9.00 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. (a) With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? (b) Where is Henrietta when she catches the bagels?

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The position of a dragonfly that is flying parallel to the ground is given as a function of time by \(\vec{r}=\left[2.90 \mathrm{m}+\left(0.0900 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\boldsymbol{u}}-\left(0.0150 \mathrm{m} / \mathrm{s}^{3}\right) t^{3} \hat{\boldsymbol{J}}\) . (a) At what value of \(t\) does the velocity vector of the insect make an angle of \(30.0^{\circ}\) clockwise from the \(+x\) -axis? (b) At the time calculated in part (a), what are the magnitude and direction of the acceleration vector of the insect?

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