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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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Most popular questions from this chapter

At the beginning of a 3.0 -h plane trip, you are traveling due north at $192 \mathrm{km} / \mathrm{h}\(. At the end, you are traveling \)240 \mathrm{km} / \mathrm{h}\( in the northwest direction \)left(45^{\circ}$ west of north). \right. (a) Draw your initial and final velocity vectors. (b) Find the change in your velocity. (c) What is your average acceleration during the trip?
A pilot wants to fly from Dallas to Oklahoma City, a distance of $330 \mathrm{km}\( at an angle of \)10.0^{\circ}$ west of north. The pilot heads directly toward Oklahoma City with an air speed of $200 \mathrm{km} / \mathrm{h}\(. After flying for \)1.0 \mathrm{h},\( the pilot finds that he is \)15 \mathrm{km}$ off course to the west of where he expected to be after one hour assuming there was no wind. (a) What is the velocity and direction of the wind? (b) In what direction should the pilot have headed his plane to fly directly to Oklahoma City without being blown off course?
Michaela is planning a trip in Ireland from Killarney to Cork to visit Blarney Castle. (See Example \(3.2 .\) ) She also wants to visit Mallow, which is located \(39 \mathrm{km}\) due east of Killarney and \(22 \mathrm{km}\) due north of Cork. Draw the displacement vectors for the trip when she travels from Killarney to Mallow to Cork. (a) What is the magnitude of her displacement once she reaches Cork? (b) How much additional distance does Michaela travel in going to Cork by way of Mallow instead of going directly from Killarney to Cork?
A projectile is launched at \(t=0\) with initial speed \(v_{\mathrm{i}}\) at an angle \(\theta\) above the horizontal. (a) What are \(v_{x}\) and \(v_{y}\) at the projectile's highest point? (b) Find the time \(t\) at which the projectile reaches its maximum height. (c) Show that the maximum height \(H\) of the projectile is $$H=\frac{\left(v_{i} \sin \theta\right)^{2}}{2 g}$$
The range \(R\) of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at \(t=0\) with initial speed \(v_{i}\) at an angle \(\theta\) above the horizontal. (a) Find the time \(t\) at which the projectile returns to its original altitude. (b) Show that the range is \(R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}\) [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of \(\theta\) gives the maximum range? What is this maximum range?
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