A bullet shot horizontally strikes the ground with a velocity of 150.0km/h (10 degrees below the horizontal), determine the height from which it was fired.
To determine the height from which the bullet was fired, we can use the equations of projectile motion. Let's break it down step by step:
1. Convert the velocity from km/h to m/s.
We need to convert the velocity of the bullet from km/h to m/s since the equations of projectile motion require velocity in meters per second.
Given: Velocity (v) = 150.0 km/h
1 kilometer = 1000 meters
1 hour = 3600 seconds
Therefore, the conversion factor is: 1 km/h = (1000/3600) m/s
Converting the velocity:
v = 150.0 km/h * (1000/3600) m/s = (150.0*1000) / (3600) m/s = 41.67 m/s (approximately)
2. Separate the horizontal and vertical components of the velocity.
Given: The bullet is shot at an angle 10 degrees below the horizontal.
The horizontal component of the velocity (Vx) remains constant throughout the motion since there are no horizontal forces acting on the bullet. Therefore, Vx = v * cos(theta), where theta is the angle of the projectile with respect to the horizontal direction.
The vertical component of the velocity (Vy) changes due to the acceleration due to gravity. Vy = v * sin(theta)
Here, theta = 10 degrees and we have v = 41.67 m/s.
Using trigonometric formulas:
Vx = 41.67 m/s * cos(10 degrees) = 41.67 * cos(10 degrees) m/s (approximately)
Vy = 41.67 m/s * sin(10 degrees) = 41.67 * sin(10 degrees) m/s (approximately)
3. Calculate the time (t) it takes for the bullet to hit the ground.
The time taken for the bullet to reach the ground is the same for both horizontal and vertical components. We can use either Vx or Vy to calculate the time.
Let's use Vy for this calculation:
Considering the vertical motion, the equation is:
h = (1/2) * g * t^2
where h is the height from which the bullet was fired and g is the acceleration due to gravity.
We know that g = 9.8 m/s^2 (approximately) and Vy = 41.67 * sin(10 degrees) m/s (approximately)
Setting h to zero (since the bullet hits the ground), we can solve for t.
0 = (1/2) * 9.8 * t^2
0 = 4.9 * t^2
Solving for t:
t^2 = 0 / 4.9
t = sqrt(0)
t = 0
We have obtained t = 0, which implies that the bullet hits the ground instantly.
4. Calculate the height (h) from which the bullet was fired.
Using either Vx or Vy, we can calculate the distance covered horizontally before hitting the ground and then determine the height from which the bullet was fired.
Using Vx:
The distance covered in the horizontal direction is given by:
d = Vx * t
Since t = 0, the distance covered in the horizontal direction is also zero.
Therefore, the height from which the bullet was fired is zero.
Conclusion:
The height from which the bullet was fired is zero, indicating it was fired from the ground level.