at a hight of 3900m,an airfore bomber maintains a constant velocity 1200km/hr toward a point directly above its target.calculate the angle of sight at which the plane should release a bomb to strick the target. take g-10m/s square.
To calculate the angle of sight at which the plane should release a bomb to strike the target, we can use trigonometry. Here's how you can solve this problem step by step:
Step 1: Identify the given information:
- Height of the bomber, h = 3900 m
- Velocity of the bomber, v = 1200 km/hr (Note: We need to convert it to m/s later)
- Acceleration due to gravity, g = -10 m/s^2 (negative since it is pointing downward)
Step 2: Convert the velocity from km/hr to m/s:
Since 1 km = 1000 m and 1 hr = 3600 s:
1200 km/hr = (1200 * 1000) m / (3600 s) = 333.33 m/s (approximately)
Step 3: Calculate the time it takes for the bomb to reach the target:
To calculate the time, we can divide the height of the bomber by the vertical component of its velocity:
time = height / vertical velocity
t = h / v
Substituting the values:
t = 3900 m / 333.33 m/s
t ≈ 11.70 s (approximately)
Step 4: Calculate the horizontal distance covered by the bomber in that time:
horizontal distance = horizontal velocity × time
The horizontal velocity is the same as the velocity of the bomber.
horizontal distance = 1200 km/hr × (3600 s / 1000 m)
horizontal distance = 1200 km/hr × 3.6 m/s (conversion factor)
horizontal distance ≈ 4320 m (approximately)
Step 5: Calculate the angle of sight:
The angle of sight at which the plane should release the bomb is the inverse tangent of the horizontal distance divided by the height of the bomber.
angle of sight = arctan(horizontal distance / height)
θ = arctan(4320 m / 3900 m)
Using a calculator, we find that the angle of sight is approximately 49.03 degrees.
Therefore, the angle at which the plane should release the bomb to strike the target is approximately 49.03 degrees.