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Chapter 3: Problem 54

After being assaulted by flying cannonballs, the knights on the castle walls ( \(12 \mathrm{m}\) above the ground) respond by propelling flaming pitch balls at their assailants. One ball lands on the ground at a distance of 50 m from the castle walls. If it was launched at an angle of \(53^{\circ}\) above the horizontal, what was its initial speed?

Short Answer

Expert verified
Answer: The initial speed of the flaming pitch ball is approximately 25.37 m/s.

Step by step solution

01

Calculate the time of flight using horizontal motion.

Since there is no air resistance, the horizontal velocity remains constant throughout the flight. Let's call the initial horizontal velocity \(v_x\) and the initial vertical velocity \(v_y\). The initial speed is \(v\). The angle of projection is given as \(53^{\circ}\). We have: \(v_{x} = v\cos(53^{\circ})\), and distance \(= v_{x} \times\) time of flight, so, \(50 = v\cos(53^{\circ}) \times\) time of flight, Let's call the time of flight \(T\). We get: \(T = \frac{50}{v\cos(53^{\circ})}\).
02

Calculate the initial speed using vertical motion.

For vertical motion, we need to consider the initial height of \(12\mathrm{m}\). The equation for the vertical motion is: \(h = v_{y}T- \frac{1}{2}gt^2\), where \(h = 12\mathrm{m}\), \(v_{y} = v\sin(53^{\circ})\), \(g = 9.81\mathrm{m/s}^2\) (acceleration due to gravity), and \(T\) is the time of flight from Step 1. Inserting the value of \(T\), we have: \(12 = v\sin(53^{\circ})\left(\frac{50}{v\cos(53^{\circ})}\right) - \frac{1}{2}\time( 9.81)\mathrm{m/s}^2\left(\frac{50}{v\cos(53^{\circ})}\right)^2\)
03

Solve for the initial speed \(v\).

Now, we solve the above equation for the initial speed \(v\): \(12 = (50\tan(53^{\circ})) - \frac{1}{2}\time( 9.81)(50)^2\frac{\sec^2(53^{\circ})}{v^2} \) From this equation, we can isolate \(v\) and find the initial speed: \(v^2 = \frac{50^2\sec^2(53^{\circ})}{12 - 50\tan(53^{\circ}) + \frac{1}{2}\time( 9.81)(50)^2\sec^2(53^{\circ})} \) Calculating using the given values, we get: \(v \approx 25.37 \mathrm{m/s}\) Therefore, the initial speed of the flaming pitch ball was approximately \(25.37 \mathrm{m/s}\).

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