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Chapter 3: Problem 68

A cannon is fired from a hill \(116.7 \mathrm{~m}\) high at an angle of \(22.7^{\circ}\) with respect to the horizontal. If the muzzle velocity is \(36.1 \mathrm{~m} / \mathrm{s}\), what is the speed of a 4.35 -kg cannonball when it hits the ground \(116.7 \mathrm{~m}\) below?

Short Answer

Expert verified
Question: Given that the initial velocity of a cannonball is 36.1 m/s at an angle of 22.7° with respect to the horizontal, and it is fired from a height of 116.7 m, determine the speed of the cannonball when it hits the ground. Answer: To determine the speed of the cannonball, first find the horizontal and vertical components of the initial velocity using trigonometric formulas. Then, calculate the time it takes for the cannonball to hit the ground using a kinematic equation. Finally, calculate the final velocities in both horizontal and vertical directions and determine the magnitude of the final velocity vector, which is the speed of the cannonball when it hits the ground.

Step by step solution

01

Find the horizontal and vertical components of the initial velocity

To find the initial velocity components, we can use the following trigonometric formulas: \\ $$ v_{0x} = v_0 \cos(\theta) $$ $$ v_{0y} = v_0 \sin(\theta) $$ where \(v_0\) is the initial velocity, \(\theta\) is the angle with respect to the horizontal, and \(v_{0x}\) and \(v_{0y}\) are the horizontal and vertical components of the initial velocity, respectively. Using the given values, we get: $$ v_{0x} = 36.1\mathrm{~m/s}\cos(22.7^\circ) $$ $$ v_{0y} = 36.1\mathrm{~m/s}\sin(22.7^\circ) $$
02

Calculate the time to hit the ground

We can use the kinematic equation to find the time it takes for the cannonball to hit the ground: $$ y = y_0 + v_{0y}t - \frac{1}{2}gt^2 $$ where \(y\) is the final vertical position, \(y_0\) is the initial vertical position, \(v_{0y}\) is the initial vertical velocity, \(t\) is the time, and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)). Since the cannonball hits the ground, \(y = 0\), and because the hill is \(116.7\ \mathrm{m}\) high, \(y_0 = -116.7\mathrm{~m}\). Now, we can solve for the time \(t\) using a quadratic equation.
03

Find the final velocities and speed

The horizontal velocity remains constant throughout the motion, so the final horizontal velocity is equal to the initial horizontal velocity: $$ v_{fx} = v_{0x} $$ The final vertical velocity can be calculated using another kinematic equation: $$ v_{fy} = v_{0y} - gt $$ The speed of the cannonball when it hits the ground can be found by calculating the magnitude of the final velocity vector, which can be computed using the Pythagorean theorem: $$ v_f =\sqrt{v_{fx}^2 + v_{fy}^2} $$ Plug in the values of \(v_{fx}\) and \(v_{fy}\) obtained from previous steps to find the speed of the cannonball when it hits the ground.

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