A bomber is flying at a velocity of 200.0 km/hr with an altitude of 1.5 km. How far (horizontally) from its intended target should the plane drop the bomb?
The bomb moves exactly as fast horizontally as the bomber if friction is ignored. (If the bomber does not turn, the bomb explodes right under it)
So how long will it take the bom to fall 1500 meters?
h = (1/2) g t^2
1500 = 4.9 t^2
t = 17.5 seconds to fall.
how far does the bomber move in 17.5 seconds?
(200,000 m/3600 s)(17.5s) = 972 meters
A launcher fires a projectile with a velocity of 6.31m/s. Determine the horizontal distance of the projectile (in the air) at 5 degrees. Recalculate this distance at 10, 15, 20.... (5 degree increments) up to 90 degrees (at which point the horizontal distance will be zero). Graph this data (Distance vs. Launch Angle). Distance will be on the Y-axis and launch angle will be on the X-axis.
To find the distance from the intended target where the bomb should be dropped, we need to consider the horizontal component of the plane's velocity.
The horizontal component of the velocity remains constant throughout the flight, assuming no external factors affect it. Therefore, we can use the formula for distance:
distance = velocity * time
In this case, we want to find the time it takes for the bomb to reach the ground. Since the altitude is given in kilometers, we need to convert it to meters for consistent units. There are 1000 meters in 1 kilometer, so the altitude is 1500 meters.
To calculate the time, we can use the equation for free fall motion:
distance = (1/2) * acceleration * time^2
In this equation, the distance is the altitude (1500 meters), the acceleration is due to gravity (9.8 m/s^2), and we want to find the time.
Substituting the given values into the equation:
1500 = (1/2) * 9.8 * time^2
Simplifying:
3000 = 9.8 * time^2
Dividing both sides by 9.8:
time^2 = 3000 / 9.8
time^2 = 306.12
Taking the square root of both sides:
time ≈ √306.12
time ≈ 17.50 seconds
Now that we have the time it takes for the bomb to reach the ground, we can find the horizontal distance using the formula mentioned earlier:
distance = velocity * time
Substituting the given values:
distance = 200 km/hr * (17.50 s / 3600 s)
Since the velocity is given in kilometers per hour and the time is in seconds, we need to convert the velocity to meters per second:
200 km/hr = 200,000 m/3600 s ≈ 55.56 m/s
Substituting the values:
distance ≈ 55.56 m/s * 17.50 s
Calculating:
distance ≈ 972 meters
Therefore, the plane should drop the bomb approximately 972 meters horizontally from its intended target.