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Chapter 3: Problem 33

A projectile is launched with an initial speed of \(60.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. The projectile lands on a hillside \(4.00\) s later. Neglect air friction. (a) What is the projectile's velocity at the highest point of its trajectory? (b) What is the straight-line distance from where the projectile was launched to where it hits its target?

Short Answer

Expert verified
The velocity at the highest point of the trajectory is \(51.96 \, m/s\) and the straight-line distance from where the projectile was launched to where it hits its target is \(207.84 \, m\).

Step by step solution

01

Find the initial velocity components

The first thing to do is to decompose the initial speed into horizontal and vertical components using trigonometry. The initial horizontal velocity \(v_{i,x}\) is given by \(v_i \times \cos(\Theta)\), and the initial vertical velocity \(v_{i,y}\) is given by \(v_i \times \sin(\Theta)\). Here, \(v_i = 60.0 \, m/s\) and \(\Theta = 30^{\circ}\). So, \(v_{i,x} = 60 \cos(30^{\circ}) = 51.96 \, m/s\), and \(v_{i,y} = 60 \sin(30^{\circ}) = 30 \, m/s\).
02

Calculate the velocity at the highest point

Now, we will solve part (a) of the problem. At the highest point in its trajectory, the projectile is only moving horizontally. This means the vertical component of the velocity is zero, but the horizontal component is unchanged. Thus, the velocity at the highest point of the trajectory is simply the initial horizontal velocity \(v_{i,x} = 51.96 \, m/s\).
03

Calculate the straight-line distance

To solve part (b), we need to find the range or horizontal distance that the projectile covers. This is simply the horizontal speed multiplied by the time, i.e., Range = \(v_{i,x} \times t\). Where \(v_{i,x} = 51.96 \, m/s\) and \(t = 4.00 \, s\). So, Range = \(51.96 \, m/s \times 4.00 \, s = 207.84 \, m\).

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Key Concepts

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

Initial Velocity Decomposition
When dealing with projectile motion, one of the first steps is to break down the initial velocity into horizontal and vertical components. This helps us understand how the projectile moves in each direction. We use basic trigonometry to do this.

For a projectile launched with an initial velocity \(v_i\) at an angle \(\Theta\), the horizontal component \(v_{i,x}\) can be found using \(v_{i,x} = v_i \times \cos(\Theta)\). The vertical component \(v_{i,y}\) is given by \(v_{i,y} = v_i \times \sin(\Theta)\).

In our example, the initial speed is \(60.0 \, m/s\) and the launch angle is \(30^{\circ}\). By applying the formulas:
  • Horizontal Velocity: \(v_{i,x} = 60 \cos(30^{\circ}) = 51.96 \, m/s\)
  • Vertical Velocity: \(v_{i,y} = 60 \sin(30^{\circ}) = 30 \, m/s\)
Breaking down the velocity this way allows us to separately analyze the horizontal and vertical motions of the projectile.
Trajectory Analysis
Understanding the trajectory of a projectile is crucial to find its highest point and other features. The trajectory is the path followed by a moving object. In projectile motion, it's usually a parabolic path.

At the highest point of the projectile's trajectory, the vertical velocity component becomes zero. This is because gravity slows the vertical speed until it momentarily stops at the top. However, the horizontal component remains constant as there is no air resistance to slow it down.

For our problem, at the highest point:
  • Vertical Velocity: 0 \, m/s
  • Horizontal Velocity: \(51.96 \, m/s\)
This helps us determine that the velocity at the highest point is purely horizontal, matching the initial horizontal component.
Horizontal Range Calculation
The horizontal range of a projectile is the distance it travels horizontally between launch and landing. To find this, we multiply the horizontal velocity by the total time the projectile is in the air.

The horizontal velocity \(v_{i,x}\) remains constant during the flight if air friction is neglected. So, the formula for range is:
  • Range = \(v_{i,x} \times t\)

In our case, with \(v_{i,x} = 51.96 \, m/s\) and flight time \(t = 4.00 \, s\), the range becomes:
  • Range: \(51.96 \, m/s \times 4.00 \, s = 207.84 \, m\)
This calculation shows how far the projectile travels horizontally from its starting point.

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Most popular questions from this chapter

A river has a steady speed of \(0.500 \mathrm{~m} / \mathrm{s}\). A student swims upstream a distance of \(1.00 \mathrm{~km}\) and swims back to the starting point. (a) If the student can swim at a speed of \(1.20 \mathrm{~m} / \mathrm{s}\) in still water, how long does the trip take? (b) How much time is required in still water for the same length swim? (c) Intuitively, why does the swim take longer when there is a current?

GP From the window of a building, a ball is tossed from a height \(y_{0}\) above the ground with an initial velocity of \(8.00 \mathrm{~m} / \mathrm{s}\) and angle of \(20.0^{\circ}\) below the horizontal. It strikes the ground \(3.00 \mathrm{~s}\) later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive \(y\)-direction, what are the initial coordinates of the ball? (b) With the positive \(x\)-direction chosen to be out the window, find the \(x\)-and \(y\)-components of the initial velocity. (c) Find the equations for the \(x\)-and \(y\)-components of the position as functions of time. (d) How far horizontally from the base of the building does the ball strike the ground? (e) Find the height from which the ball was thrown. (f) How long does it take the ball to reach a point \(10.0 \mathrm{~m}\) below the level of launching?

QIC A Nordic jumper goes off a ski jump at an angle of \(10.0^{\circ}\) below the horizontal, traveling \(108 \mathrm{~m}\) horizontally and \(55.0 \mathrm{~m}\) vertically before landing. (a) Ignoring friction and aerodynamic effects, calculate the speed needed by the skier on leaving the ramp. (b) Olympic Nordic jumpers can make such jumps with a jump speed of \(23.0 \mathrm{~m} / \mathrm{s}\), which is considerably less than the answer found in part (a). Explain how that is possible.

S In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strikes the floor at distance \(d\) from the base of the counter. If the height of the counter is \(h\), (a) find an expression for the time \(t\) it takes the cup to fall to the floor in terms of the variables \(h\) and \(g\). (b) With what speed does the mug leave the counter? Answer in terms of the variables \(d, g\), and \(h .\) (c) In the same terms, what is the speed of the cup immediately before it hits the floor? (d) In terms of \(h\) and \(d\), what is the direction of the cup's velocity immediately before it hits the floor?

BIO A chinook salmon has a maximum underwater speed of \(3.58 \mathrm{~m} / \mathrm{s}\), but it can jump out of water with a speed of \(6.26 \mathrm{~m} / \mathrm{s}\). To move upstream past a waterfall, the salmon does not need to jump to the top of the fall, but only to a point in the fall where the water speed is less than \(3.58 \mathrm{~m} / \mathrm{s} ;\) it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, let's assume it can swim to the top if the water speed is \(3.00 \mathrm{~m} / \mathrm{s}\). If water has a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) as it passes over a ledge, (a) how far below the ledge will the water be moving with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) ? (Note that water undergoes projectile motion once it leaves the ledge.) (b) If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?
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