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Chapter 3: Problem 93

A beanbag is thrown horizontally from a dorm room window a height \(h\) above the ground. It hits the ground a horizontal distance \(h\) (the same distance \(h\) ) from the dorm directly below the window from which it was thrown. Ignoring air resistance, find the direction of the beanbag's velocity just before impact.

Short Answer

Expert verified
Answer: The direction of the beanbag's velocity just before impact is given by the angle, \(\theta\), which can be calculated using the following equation: $$ \theta = \tan^{-1}\left(\frac{\sqrt{2gh}}{\sqrt{\frac{gh}{2}}}\right) $$ Here, \(g\) is the gravitational acceleration (approximately \(9.81\,\text{m/s}^2\)), and \(h\) is the height from which the beanbag is thrown.

Step by step solution

01

Identify the given information

The beanbag is thrown horizontally from a height, \(h\), above the ground. It hits the ground a horizontal distance, \(h\) from the dorm directly below the window from which it was thrown. We need to find the direction of the beanbag's velocity just before impact.
02

Apply the equations of motion

First, let's consider the horizontal motion of the beanbag. Since there is no air resistance, the horizontal velocity, \(v_x\), remains constant. Now, let's look at the vertical motion of the beanbag. It is initially at rest and then falls freely under the influence of gravity. We can use the following equation for the vertical motion: $$ y = h - \frac{1}{2}gt^2 $$ Where \(y\) is the distance traveled in the vertical direction, \(g\) is the gravitational acceleration (approximately \(9.81\,\text{m/s}^2\)), and \(t\) is the time taken for the beanbag to travel from the window to the ground.
03

Find the time of flight

First, we need to find the time it takes for the beanbag to hit the ground. We can use the vertical motion equation and solve for \(t\): $$ h = \frac{1}{2}gt^2 $$ Rearrange the equation to find the time, \(t\): $$ t = \sqrt{\frac{2h}{g}} $$
04

Determine the horizontal velocity

Now that we have the time of flight, we can find the horizontal velocity, \(v_x\). The horizontal distance is equal to the product of the horizontal velocity and the time of flight: $$ h = v_x t $$ Substituting the time of flight equation from Step 3: $$ h = v_x \sqrt{\frac{2h}{g}} $$ Now, solve for the horizontal velocity, \(v_x\): $$ v_x = \sqrt{\frac{gh}{2}} $$
05

Calculate the vertical velocity just before impact

Next, we need to find the vertical velocity, \(v_y\), just before the beanbag hits the ground. We can use the following equation for the vertical velocity: $$ v_y = gt $$ Substituting the time of flight equation from Step 3: $$ v_y = g\sqrt{\frac{2h}{g}} $$ $$ v_y = \sqrt{2gh} $$
06

Determine the direction of the velocity just before impact

Now, we can find the direction of the velocity just before impact by calculating the angle, \(\theta\), between the horizontal and the velocity vector using the horizontal and vertical velocities: $$ \tan\theta = \frac{v_y}{v_x} $$ Substituting the values of \(v_x\) and \(v_y\) calculated in steps 4 and 5: $$ \tan\theta = \frac{\sqrt{2gh}}{\sqrt{\frac{gh}{2}}} $$ Now, finding the value of the angle, \(\theta\): $$ \theta = \tan^{-1}\left(\frac{\sqrt{2gh}}{\sqrt{\frac{gh}{2}}}\right) $$ The angle, \(\theta\), represents the direction of the beanbag's velocity just before impact.

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