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Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree that is \(4.0 \mathrm{m}\) above the gun. The base of the tree is \(20 \mathrm{m}\) away. The speed of the paintball as it leaves the gun is \(50 \mathrm{m} / \mathrm{s}\). How far below the knothole does the paintball strike the tree?

Short Answer

Expert verified
The paintball will strike the tree \(0.7848 \, \mathrm{m}\) below the knothole.

Step by step solution

01

Calculate the time of flight

The first step is to calculate the time it takes for the paintball to hit the tree. Because the distance traveled horizontally, \(d\), is given to be 20 meters, and we know the horizontal speed, \(v\), is 50 m/s, we can use the equation \(d = v \cdot t\) to solve for time, \(t = d/v = 20 \, \mathrm{m} / 50 \, \mathrm{m/s} = 0.4 \, \mathrm{s}\). This is the time it takes the paintball to reach the tree, which we will use in the next step.
02

Calculate final vertical position

Using the equation of motion \(h = v_{0i}t + 0.5 \cdot g \cdot t^{2}\) where initial vertical velocity \(v_{0i}\) is zero (since forester aims gun directly at knothole, meaning no initial vertical speed), while \(g\) is acceleration due to gravity, -9.81 m/s², and \(t\) is 0.4 s. Inserting these values gives us the final height, \(h = -0.5 * 9.81 * (0.4)^2 = -0.7848\, \mathrm{m}\). \(h\) is the distance fallen away from the initial height of trajectory as the pellet was initially fired directly at the knothole.
03

Calculate the total vertical displacement

Since the initial height was 4.0 meters and the displacement, \(h\) below this point is calculated to be -0.7848 meters, we add these together to find that the paintball hits the tree at \(4.0 - 0.7848 = 3.2152 \, \mathrm{m}\). This means the ball lands \(4.0 - 3.2152 = 0.7848 \, \mathrm{m}\) below the initial height of 4.0 m.

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Key Concepts

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

Kinematics in Physics
Kinematics is a branch of physics that deals with the description of motion without considering its causes. It involves analysing the positions, velocities, and accelerations of objects and how these quantities are related through time.

In our paintball scenario, the motion of the paintball can be dissected into two components: horizontal and vertical. The horizontal motion is uniform because there are no horizontal forces acting on the paintball once it leaves the gun, meaning it has a constant horizontal velocity. The vertical motion, however, is influenced by gravity, resulting in a uniformly accelerated movement.

Breaking Down Motion

Kinematics allows us to break down the paintball's flight into two independent movements and study them with equations of motion. For horizontal displacement, we use simple formulas that pertain to uniform motion. For vertical displacement, we bring into play additional kinematics equations that factor in acceleration to describe the effects of gravity, which is vital for obtaining the point of impact of the paintball.
Time of Flight Calculation
The time of flight in projectile motion refers to the total time an object remains in the air. Calculating this is essential for predicting where and when the object will land.

In our exercise, the time of flight is determined by the horizontal component of motion since the horizontal speed remains constant and no other forces are acting on the paintball in that direction. The basic equation used here is distance equals speed multiplied by time, which is rearranged to solve for time.

Formula Application

Once we've found the time, we've identified a 'link' between the horizontal and vertical components - they occur over the same amount of time. This concept is crucial for understanding that the paintball spends the same duration moving horizontally to cover the distance to the tree as it does falling vertically under the force of gravity.
Horizontal and Vertical Displacement
Displacement refers to the change in position of an object and is a vector quantity, meaning it has both magnitude and direction. In projectile motion, an object moves in both horizontal and vertical directions, creating a trajectory that can be analyzed separately in each direction.

Horizontal displacement occurs at a consistent speed when there are no opposing forces, like air resistance, which is normally negligible in textbook problems. On the other hand, vertical displacement involves dealing with gravity, which accelerates the object downwards.

Understanding Combined Displacement

By calculating both horizontal and vertical displacements independently, we can understand the position of the projectile at any point in its trajectory. The vertical displacement, in particular, informs how far below the intended target the object lands, as seen in our exercise where the paintball falls short of the knothole after traveling through the air.

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Most popular questions from this chapter

Ships A and B leave port together. For the next two hours, ship \(\mathrm{A}\) travels at \(20 \mathrm{mph}\) in a direction \(30^{\circ}\) west of north while ship \(\mathrm{B}\) travels \(20^{\circ}\) east of north at \(25 \mathrm{mph}\). a. What is the distance between the two ships two hours after they depart? b. What is the speed of ship \(A\) as seen by ship \(B ?\)

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Each of the following vectors is given in terms of its \(x\) - and \(y\) -components. Draw the vector, label an angle that specifies the vector's direction, then find the vector's magnitude and direction. a. \(v_{x}=20 \mathrm{m} / \mathrm{s}, v_{y}=40 \mathrm{m} / \mathrm{s}\) b. \(a_{x}=2.0 \mathrm{m} / \mathrm{s}^{2}, a_{y}=-6.0 \mathrm{m} / \mathrm{s}^{2}\)

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