**Projectile Motion Problem**A bullet is fired through a board \(20.0\) cm thick in such a way that the bullet's line of motion is perpendicular to the face of the board. The initial speed of the bullet is \(530\) m/s and it emerges from the other side of the board with a speed of \(430\) m/s.1. **(a) Find the acceleration of the bullet as it passes through the board.** \[\_\_\_\_\_\_\_\_\_\_\_\_\_\_] \(\text{m/s}^2\]2. **(b) Find the total time the bullet is in contact with the board.** \[\_\_\_\_\_\_\_\_\_\_\_\_\_\_] \(\text{s}\]---For educational purposes:The problem presented involves a bullet moving through a medium (a board), experiencing deceleration due to the resistance encountered within the medium. The initial and final velocities are provided, along with the thickness of the board, enabling the calculation of both the acceleration (negative, as the bullet is slowing down) and the time of contact with the board through the application of kinematic equations.**Steps to solve:**1. **Determine the acceleration \( a \):**Using the kinematic equation:\[ v^2 = u^2 + 2as \]where,\( u = 530 \text{ m/s} \) (initial velocity),\( v = 430 \text{ m/s} \) (final velocity),\( s = 0.20 \text{ m} \) (thickness of the board, converted to meters).Rearrange to solve for \( a \):\[ a = \frac{v^2 - u^2}{2s} \]2. **Determine the time \( t \) the bullet is in contact with the board:**Using the kinematic equation:\[ v = u + at \]Rearrange to solve for \( t \):\[ t = \frac{v - u}{a} \]These equations provide a systematic method to solve for the acceleration and time of contact for the bullet in a uniform motion problem involving resistance.

Name: **Projectile Motion Problem**A bullet is fired through a board \(20.0
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: **Projectile Motion Problem**A bullet is fired through a board \(20.0\) cm thick in such a way that the bullet's line of motion is perpendicular to the face of the board. The initial speed of the bullet is \(530\) m/s and it emerges from the other s