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Chapter 3: Problem 52

A cannon, located \(60.0 \mathrm{~m}\) from the base of a vertical \(25.0-\mathrm{m}-\)tall cliff, shoots a \(15 \mathrm{~kg}\) 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 \(25.0 \mathrm{~m}\) above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

Short Answer

Expert verified
In order to clear the cliff, the minimum muzzle velocity of the shell should be calculated from the equation derived in step 2. The horizontal distance to the place where the shell lands past the cliff can be calculated using the total time in the air and initial velocity.

Step by step solution

01

Clarify the Variables

Given: Launch Angle \(\Theta = 43.0^\circ\), base to cliff distance \(X = 60.0m\), cliff height \(f = 25.0m\). To find: Initial (Muzzle) velocity and distance travelled by shell to land past the edge of the cliff.
02

Equation of projectile motion

The initial velocity minimum \(v_0\) required for the shell to clear the cliff can be calculated by resolving the motion into two independent one-dimensional motions. The upward motion under constant acceleration can be described by the equation \[f = v_0sin(\Theta)\Delta t - 1/2gt^2\], where \(f\) is the cliff height, \(t\) is the time taken, \(g\) is the acceleration due to gravity, and \(\Theta\) is the launch angle. We also know that the horizontal distance travelled by the shell can be given by \(X = v_0cos(\Theta)\Delta t\). From above equation, we can express \(\Delta t\) as \(X/ v_0cos(\Theta)\) and by substituting this in the first equation, we get the relation between \(v_0\), \(\Theta\), \(f\), \(X\) and \(g\).
03

Calculate muzzle velocity

Now substituting the given values in the equation and solving it for \(v_0\), we get the minimum muzzle velocity required for the shell to clear the cliff.
04

Calculate landing distance

For part (b), the total time in the air can be found by adding the time for shell to reach the cliff with the time for it to fall from the height of the cliff to the ground. The total distance it traveled horizontally can be calculated as \(v_0Tcos(\Theta)\), where \(T\) is the total time in the air. Calculating this will give us the horizontal distance shell lands past the edge of the cliff.

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Key Concepts

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

Muzzle Velocity
Muzzle velocity is a term used to describe the speed at which a projectile, such as a cannonball or bullet, is launched from a firearm. This velocity determines how fast and how far the projectile will travel once it's out of the cannon.
In the context of this problem, the muzzle velocity must be sufficient for the cannonball to just clear the top of a cliff that's 25 meters tall.
To find this minimum speed, we break down the projectile's motion into horizontal and vertical components.
Here's why muzzle velocity is crucial:
  • **Initial Speed**: It's the starting speed of the projectile and has to overcome gravitational pull to reach a certain height, in this case, the height of the cliff.
  • **Launch Angle Impact**: Without a sufficient speed at a given angle, the object won't travel the planned distance or reach the desired height.
When solving the problem, knowing the launch angle (43 degrees) allows us to use equations of motion to calculate the required muzzle velocity. This can be seen as similar to breaking down a force into components that act vertically and horizontally, making it easier to understand how each part affects the overall trajectory.
Horizontal Motion
When we talk about projectile motion, the horizontal component is independent of the vertical motion.
This means that horizontal motion happens with a constant velocity because there's no acceleration in the horizontal direction (assuming air resistance is negligible).
For a shell moving horizontally:
  • **Constant Velocity**: Once launched, the horizontal speed remains constant throughout the flight, defined by the formula: \[ v_0 \cos(\Theta) \]. Here, \( v_0 \) is the muzzle velocity and \( \Theta \) is the launch angle (43 degrees in this problem).
  • **Distance Covered**: The horizontal distance depends on this velocity and the total time the projectile is in the air, calculated as \( v_0T\cos(\Theta) \).
In our problem, horizontal motion helps determine how far the shell can travel; importantly, whether it reaches beyond the cliff after clearing it.
This understanding enables estimation of how far the shell lands past the edge of the cliff once it flies over.
Vertical Motion
Vertical motion in projectile motion is influenced by gravity. This motion is more complex as it involves changing velocities and acceleration due to gravity.
Here's what guides the vertical component:
  • **Initial Vertical Velocity**: Calculated using \( v_0 \sin(\Theta) \), it defines how high and how fast the projectile rises initially.
  • **Acceleration due to Gravity**: Constantly keeps pulling the projectile downward at approximately 9.8 m/s虏. This affects both the time the projectile is airborne and its maximum height.
  • **Cliff Height Relevance**: For the shell to clear the cliff, its vertical motion must offset the cliff鈥檚 25-meter height.
In the exercise, we calculate the time taken to reach the vertical height of the cliff and then use this time to ensure the projectile also travels the horizontal distance of 60 meters.
By correctly balancing both vertical and horizontal components, we ascertain the necessary initial conditions for the projectile to succeed in clearing the cliff and beyond.

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Most popular questions from this chapter

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by \(\overrightarrow{\boldsymbol{v}}=\) \(\left[5.00 \mathrm{~m} / \mathrm{s}-\left(0.0180 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\right] \hat{\imath}+\left[2.00 \mathrm{~m} / \mathrm{s}+\left(0.550 \mathrm{~m} / \mathrm{s}^{2}\right) t\right] \hat{\jmath}\) (a) What are \(a_{x}(t)\) and \(a_{y}(t),\) the \(x\) - and \(y\) -components of the car's velocity as functions of time? (b) What are the magnitude and direction of the car's velocity at \(t=8.00 \mathrm{~s} ?\) (b) What are the magnitude and direction of the car's acceleration at \(t=8.00 \mathrm{~s} ?\)

The position of a squirrel running in a park is given by \(\vec{r}=\left[(0.280 \mathrm{~m} / \mathrm{s}) t+\left(0.0360 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\imath}+\left(0.0190 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \hat{\jmath} .\) (a) What are \(v_{x}(t)\) and \(v_{y}(t),\) the \(x\) - and \(y\) -components of the velocity of the squirrel, as functions of time? (b) At \(t=5.00 \mathrm{~s}\), how far is the squirrel from its initial position? (c) At \(t=5.00 \mathrm{~s}\), what are the magnitude and direction of the squirrel's velocity?

An airplane pilot sets a compass course due west and maintains an airspeed of \(220 \mathrm{~km} / \mathrm{h}\). After flying for \(0.500 \mathrm{~h},\) she finds herself over a town \(120 \mathrm{~km}\) west and \(20 \mathrm{~km}\) south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is \(40 \mathrm{~km} / \mathrm{h}\) due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of \(220 \mathrm{~km} / \mathrm{h}\).

A squirrel has \(x\) - and \(y\) -coordinates \((1.1 \mathrm{~m}, 3.4 \mathrm{~m})\) at time \(t_{1}=0\) and coordinates \((5.3 \mathrm{~m},-0.5 \mathrm{~m})\) at time \(t_{2}=3.0 \mathrm{~s}\). For this time interval, find (a) the components of the average velocity, and (b) the magnitude and direction of the average velocity.

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