/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A rock is thrown with a velocity... [FREE SOLUTION] | 91Ó°ÊÓ

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A rock is thrown with a velocity \(v_0\), at an angle of \(\alpha_0\) from the horizontal, from the roof of a building of height \(h\). Ignore air resistance. Calculate the speed of the rock just before it strikes the ground, and show that this speed is independent of \(\alpha_0\).

Short Answer

Expert verified
The speed of the rock just before it strikes the ground is \(\sqrt{v_0^2 + 2gh}\), independent of \(\alpha_0\).

Step by step solution

01

Resolve Initial Velocity

First, decompose the initial velocity \(v_0\) into horizontal and vertical components. The horizontal component is \(v_{0x} = v_0 \cos(\alpha_0)\) and the vertical component is \(v_{0y} = v_0 \sin(\alpha_0)\).
02

Determine Time of Flight

Use the vertical motion to find the time it takes for the rock to hit the ground. The equation of motion for vertical displacement is \(h + v_{0y}t + \frac{1}{2}g t^2 = 0\). Solve this quadratic equation for \(t\).
03

Calculate Final Vertical Velocity

At the moment the rock strikes the ground, its vertical velocity can be calculated using \(v_y = v_{0y} + g t\), where \(g\) is the acceleration due to gravity.
04

Determine Final Horizontal Velocity

The horizontal velocity remains constant during the flight, so \(v_x = v_{0x}\).
05

Calculate the Magnitude of the Final Velocity

The speed of the rock just before it strikes the ground is the magnitude of its final velocity vector: \(v_f = \sqrt{v_x^2 + v_y^2}\).
06

Show Independence from \(\alpha_0\)

Substitute the expressions for \(v_x\) and \(v_y\) into the formula for \(v_f\). Simplify to show that \(v_f = \sqrt{v_0^2 + 2gh}\), demonstrating that the speed is independent of \(\alpha_0\).

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Key Concepts

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

Kinematics in Projectile Motion
Kinematics is a branch of physics that deals with the motion of objects without considering the causes of motion. In projectile motion, we're interested in the trajectory of an object that is thrown or projected into the air. This type of motion can be analyzed by breaking it down into horizontal and vertical components.
The key variables involved include displacement, velocity, acceleration, and time. When analyzing projectile motion, kinematics provides us with the equations of motion, which help us determine various parameters like the time of flight, range, and final velocity.
Here are some important aspects to consider:
  • **Displacement:** The overall change in position of the projectile from its starting point to its ending point.
  • **Velocity:** This includes both the initial velocity and the velocity components in horizontal and vertical directions.
  • **Acceleration:** In the vertical direction, the acceleration is usually due to gravity, while in horizontal motion, it's constant if air resistance is neglected.
  • **Time:** It's crucial to know how long the projectile is in motion, which can often be found by referring to vertical displacement equations.
These kinematic principles are foundational to solving problems involving projectiles, letting us predict behavior and results of the motion had.
Understanding Initial Velocity Components
Initial velocity components are crucial in analyzing projectile motion. When a projectile is launched at an angle, its initial velocity can be split into two separate components: horizontal and vertical. By decomposing the velocity, we can treat these components independently, which simplifies calculations.
Here's how this works:
  • **Horizontal Velocity Component:** Represented as \(v_{0x} = v_0 \cos(\alpha_0)\). This is the portion of the velocity responsible for moving the projectile horizontally. It remains constant throughout the journey because we ignore air resistance.
  • **Vertical Velocity Component:** Represented as \(v_{0y} = v_0 \sin(\alpha_0)\). This part affects how high the projectile will rise or how it will descend. It's influenced by gravity, which changes its magnitude over time.
By analyzing these components separately, we can more easily find out how long a projectile moves, how far, and what path it takes, culminating in understanding its final motion outcomes.
Final Velocity in Projectile Motion
The final velocity of a projectile is a combined effect of its horizontal and vertical components just before it hits the ground. To find this, we use the following kinematic principles and equations:
  • **Vertical Final Velocity:** Since the vertical component of velocity changes due to gravity, just before impact it is given by \(v_y = v_{0y} + g t\), where \(g\) is the gravitational acceleration and \(t\) is the time of flight.
  • **Horizontal Final Velocity:** The horizontal component does not change during the flight, so \(v_x = v_{0x}\).
  • **Magnitude of Final Velocity:** The overall speed of the projectile just before it hits the ground is the magnitude of the vector, \(v_f = \sqrt{v_x^2 + v_y^2}\).
Here, we sum the squares of the horizontal and vertical components and take the square root to find the magnitude. This gives us the final velocity, a key measure of how fast the projectile travels upon impact.
Independence from Launch Angle
One intriguing aspect of projectile motion is that, under certain conditions, the speed of the object just before hitting the ground is independent of its launch angle. This principle arises from the way initial velocities decompose and evolve over time.
When calculating the final velocity, substituting the horizontal and vertical components into the equation \(v_f = \sqrt{v_x^2 + v_y^2}\) and simplifying, we find that the expression for speed simplifies to \(v_f = \sqrt{v_0^2 + 2gh}\).
This formula shows that the speed just before impact depends only on the initial speed \(v_0\) and the height \(h\) from which it is projected, not the angle \(\alpha_0\). Two outcomes underpin this:
  • The horizontal component remains constant and merely reflects how far the object travels.
  • The time of flight and the peak height depend on the angle, but the final speed depends on combined energy principles rather than angle specifics.
Understanding this independence helps appreciate energy conservation dynamics in projectile motion.

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Most popular questions from this chapter

According to \(Guinness\) \(World\) \(Records\), the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor league game. The ball traveled 188 m (618 ft) before landing on the ground outside the ballpark. (a) If the ball's initial velocity was in a direction 45\(^{\circ}\) above the horizontal, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 m (3.0 ft) above ground level? Ignore air resistance, and assume that the ground was perfectly flat. (b) How far would the ball be above a fence 3.0 m (10 ft) high if the fence was 116 m (380 ft) from home plate?

Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance \(D\) from the foot of the table. This starship now lands on the unexplored Planet \(X\). The commander, Captain Curious, rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance 2.76\(D\) from the foot of the table. What is the acceleration due to gravity on Planet \(X\)?

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9\(^\circ\) above the horizontal. Ignore air resistance. (a) At what \(two\) times is the baseball at a height of 10.0 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a). (c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?

A rhinoceros is at the origin of coordinates at time \(t_1\) = 0. For the time interval from \(t_1\) = 0 to \(t_2\) = 12.0 s, the rhino's average velocity has \(x\)-component -3.8 m/s and y-component 4.9 m/s. At time \(t_2\) = 12.0 s, (a) what are the \(x\)- and \(y\)-coordinates of the rhino? (b) How far is the rhino from the origin?

A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0\(^{\circ}\) above the horizontal toward the cliff. (a) What must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

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