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Chapter 3: Problem 58

In an arcade game, a ball is launched from the corner of a smooth inclined plane. The inclined plane makes a \(30.0^{\circ}\) angle with the horizontal and has a width of \(w=50.0 \mathrm{~cm}\) The spring-loaded launcher makes an angle of \(45.0^{\circ}\) with the lower edge of the inclined plane. The goal is to get the ball into a small hole at the opposite corner of the inclined plane. With what initial velocity should you launch the ball to achieve this goal?

Short Answer

Expert verified
The given inclined plane has dimensions of \(50.0 \mathrm{~cm}\) width and an inclination angle of \(30^\circ\). A ball is launched from one corner with a launch angle of \(45^\circ\) with the lower edge. Calculate the initial velocity needed for the ball to land in the hole at the opposite corner of the inclined plane.

Step by step solution

01

Determine the horizontal and vertical distances

Given the dimensions of the inclined plane, we need to find the horizontal and vertical distances to the hole from the launching point. The width of the inclined plane is \(50.0 \mathrm{~cm}\), so the horizontal distance \(x\) and the vertical distance \(y\) to the hole can be found using the trigonometric functions for triangles: \(x = w \sin(\theta) = 50 \sin(30^\circ)\) \(y = w \cos(\theta) = 50 \cos(30^\circ)\) where \(\theta = 30^\circ\). Calculate \(x\) and \(y\).
02

Find the initial horizontal and vertical velocities

Now we need to find the initial horizontal \(v_{0x}\) and vertical velocities \(v_{0y}\) of the ball. As the launcher makes a \(45^\circ\) angle with the lower edge of the inclined plane, we can calculate the initial velocities using: \(v_{0x} = v_0 \cos(45^\circ - \theta)\) \(v_{0y} = v_0 \sin(45^\circ - \theta)\) where \(v_0\) is the initial velocity of the ball, which we are trying to find, and \(\theta = 30^\circ\).
03

Find the time taken by ball to reach the destination

To find the time taken for the ball to reach the hole, we can consider the horizontal motion. As the horizontal acceleration \(a_x = 0\), the ball will cover the horizontal distance \(x\) with uniform velocity \(v_{0x}\). We can write this as: \(t = \frac{x}{v_{0x}}\).
04

Set up the equation for the vertical motion and solve for \(v_0\)

We will now consider the vertical motion of the ball. The equation for the vertical motion can be written as: \(y = v_{0y}t - \frac{1}{2} g t^2\) where \(g\) is the acceleration due to gravity. Substituting the expressions for \(t\) and \(v_{0y}\) from the previous step, and using the fact that \(v_{0x} = v_0 \cos(\phi)\), where \(\phi = 45^\circ - \theta\), we get: \(y = v_0 \sin(\phi) \frac{x}{v_0 \cos(\phi)} - \frac{1}{2} g \left(\frac{x}{v_0 \cos(\phi)}\right)^2\). Simplify the equation and solve for \(v_0\). This will give the initial velocity required to launch the ball into the hole at the opposite corner of the inclined plane.

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