Certainly! Let's solve this problem step-by-step.
Step 1: Understand the problem
We have two paper screens A and B separated by a horizontal distance of 150 meters. The bullet pierces through both screens, and the hole in screen B is 15 centimeters (0.15 meters) below the hole in screen A. We need to find the velocity of the bullet at the time it pierces screen A. We are given the acceleration due to gravity, [tex]\( g = 10 \, \text{m/s}^2 \)[/tex].
Step 2: Determine the time it takes for the bullet to fall 0.15 meters
Since the bullet is initially traveling horizontally, it has no initial vertical velocity component. The bullet falls purely under the influence of gravity.
Using the equation for vertical motion under constant acceleration:
[tex]\[ d = \frac{1}{2} g t^2 \][/tex]
where:
- [tex]\( d \)[/tex] is the vertical distance (0.15 meters)
- [tex]\( g \)[/tex] is the acceleration due to gravity (10 meters per second squared)
- [tex]\( t \)[/tex] is the time in seconds
Rearranging to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \sqrt{\frac{2d}{g}} \][/tex]
Step 3: Calculate the time [tex]\( t \)[/tex] (already determined)
[tex]\[ t = 0.17320508075688773 \, \text{seconds} \][/tex]
Step 4: Determine the horizontal velocity of the bullet
The bullet travels a horizontal distance of 150 meters in the time [tex]\( t \)[/tex]. The horizontal velocity [tex]\( v \)[/tex] can be determined using the relationship:
[tex]\[ v = \frac{d_{\text{horizontal}}}{t} \][/tex]
Substituting the known values:
[tex]\[ v = \frac{150 \, \text{meters}}{0.17320508075688773 \, \text{seconds}} \][/tex]
Step 5: Calculate the horizontal velocity [tex]\( v \)[/tex] (already determined)
[tex]\[ v = 866.0254037844386 \, \text{m/s} \][/tex]
Conclusion
The velocity of the bullet at the time it pierces screen A is approximately [tex]\( 866.03 \, \text{m/s} \)[/tex].
