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Chapter 6: Problem 33

Interactive Solution 6.33 at presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a speed of \(14.0 \mathrm{~m} / \mathrm{s}\). The building is \(31.0 \mathrm{~m}\) tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

Short Answer

Expert verified
(a) 28.35 m/s, (b) 28.36 m/s, (c) 28.35 m/s.

Step by step solution

01

Understand the Problem

We need to find the speed when the pebble hits the ground from different angles: horizontally, vertically up, and vertically down. The initial speed is 14.0 m/s and the height is 31.0 m. We'll use free fall dynamics.
02

Horizontal Projection (a)

When fired horizontally, the initial vertical speed is zero. Use the formula for the final velocity: \( v^2 = u^2 + 2gh \) where \( u = 0 \), \( g = 9.8\, \text{m/s}^2 \), \( h = 31\, \text{m} \). Substitute to get \( v^2 = 2 \times 9.8 \times 31 \). Calculate \( v \) and combine with horizontal speed using \( v_{total} = \sqrt{(v_{horizontal})^2 + (v_{vertical})^2} \).
03

Calculate Horizontal Projection Speed

\( v = \sqrt{2 \times 9.8 \times 31} = \sqrt{607.6} = 24.64 \text{ m/s} \). Total speed is \( \sqrt{(14.0)^2 + (24.64)^2} = \sqrt{196 + 607.6} = \sqrt{803.6} \approx 28.35 \text{ m/s} \).
04

Vertical Projection - Upwards (b)

When fired vertically up, use energy conservation. Initial kinetic energy converts into potential energy at the top, plus kinetic again when falling: \( v = \sqrt{u^2 + 2gh} \) where \( u = 14.0 \text{ m/s} \), \( g = 9.8 \text{ m/s}^2 \), \( h = 31 + \) additional height from vertical projection. Calculate maximum height and total fall.
05

Calculate Vertical Upward Speed

Use \( u^2/2g\) to find the additional height reached, \( \text{height}_{up} = 10 \text{ m} \). Total fall height = 41 m. Final speed \( \sqrt{0 + 2 \times 9.8 \times 41} = 28.36 \text{ m/s} \).
06

Vertical Projection - Downwards (c)

When fired vertically downwards, initial speed used straight in: \( v = \sqrt{u^2 + 2gh} = \sqrt{(14.0)^2 + 2 \times 9.8 \times 31} \).
07

Calculate Downward Speed

Calculate \( v = \sqrt{196 + 607.6} = \sqrt{803.6} \approx 28.35 \text{ m/s} \).
08

Summary of Results

(a) Horizontal: 28.35 m/s. (b) Upward: 28.36 m/s. (c) Downward: 28.35 m/s.

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Key Concepts

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

Free Fall Dynamics
Free fall dynamics involves objects moving solely under the influence of gravity. When the pebble is fired from the building, its vertical motion is dictated by gravitational acceleration, which is a constant at approximately 9.8 m/s虏. This means, irrespective of other motions (like horizontal), the pebble will accelerate downwards at this rate. Free fall dynamics help us determine how fast and how far objects will fall in such situations. This understanding is crucial in predicting impact speeds, particularly when air resistance is ignored, as is common in basic physics problems.
Kinematic Equations
Kinematic equations are a set of formulas used to describe the motion of objects. They account for initial velocity, acceleration, time, and displacement. For projectile motion problems like the pebble from the building, kinematic equations allow us to calculate how fast an object will be moving before it hits the ground. For example, the equation \( v^2 = u^2 + 2gh \) combines initial velocity \( u \), acceleration due to gravity \( g \), and height \( h \) to find the final velocity \( v \). This helps in solving parts of the problem where the pebble is projected vertically, either upwards or downwards.
Conservation of Energy
The principle of conservation of energy states that the total energy in a closed system remains constant. For the pebble fired upwards, its initial kinetic energy converts into potential energy as it rises, then back to kinetic as it descends. We can express this as: the kinetic energy at launch plus potential energy at the top equals kinetic energy just before impact. Hence, using the energy conservation approach simplifies calculations, affirming that initial kinetic energy is regained as potential energy is lost during descent. This reinforces the usefulness of energy principles in calculating projectile motions.
Initial Velocity
Initial velocity is the speed and direction an object has when it begins its motion. In projectile problems, initial velocity significantly affects the trajectory and final speed. For the pebble, this initial velocity was 14.0 m/s.
  • Horizontally projected, it maintains this speed until impact.
  • Vertically upwards, it first reduces to zero at its highest point before gravity accelerates it downwards.
  • Downwards, it combines with gravity for a higher impact speed.
Understanding initial velocity helps predict the motion path and energy distribution throughout the object's trajectory.

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Most popular questions from this chapter

A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the \(+x\) axis. As a result, the kinetic energy of the sled increases by \(38 \% .\) By what percentage would the sled's kinetic energy have increased if this force had pointed \(62^{\circ}\) above the \(+x\) axis?

A truck is traveling at \(11.1 \mathrm{~m} / \mathrm{s}\) down a hill when the brakes on all four wheels lock. The hill makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is \(0.750\). How far does the truck skid before coming to a stop?

Multiple-Concept Example 5 reviews many of the concepts that play a role in this problem. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of \(25.0^{\circ}\) with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200 . She coasts down a distance of \(10.4 \mathrm{~m}\) before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is \(3.50 \mathrm{~m}\) below the edge. How fast is she going just before she lands?

The motor of a ski boat generates an average power of \(7.50 \times 10^{4} \mathrm{~W}\) when the boat is moving at a constant speed of \(12 \mathrm{~m} / \mathrm{s}\). When the boat is pulling a skier at the same speed, the engine must generate an average power of \(8.30 \times 10^{4} \mathrm{~W}\). What is the tension in the tow rope that is pulling the skier?

A \(3.00-\mathrm{kg}\) model rocket is launched straight up. It reaches a maximum height of \(1.00 \times 10^{2} \mathrm{~m}\) above where its engine cuts out, even though air resistance performs \(-8.00 \times 10^{2} \mathrm{~J}\) of work on the rocket. What would have been this height if there were no air resistance?
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