An aeroplane moving horizontally at 150m/s releases a bomb at a hight of 150m.The bomb hits the intended target.What is the horizontal distance of the plane from the target when the bomb was released?
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To determine the horizontal distance of the plane from the target when the bomb was released, we need to consider the time it takes for the bomb to reach the ground.
First, we need to find the time it takes for the bomb to fall to the ground. We can use the equation for vertical motion:
h = (1/2) * g * t^2
Where:
h = vertical distance (150 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation, we get:
t^2 = (2h) / g
t^2 = (2 * 150) / 9.8
t^2 = 30.6122
t ≈ √30.6122
t ≈ 5.528 s
Now, we can use the horizontal distance equation:
d = v * t
Where:
d = horizontal distance
v = horizontal velocity of the plane (150 m/s)
t = time (5.528 s)
Plugging in the values, we get:
d = 150 * 5.528
d ≈ 829.2 m
Therefore, the horizontal distance of the plane from the target when the bomb was released is approximately 829.2 meters.
To find the horizontal distance of the plane from the target when the bomb was released, we can use the concept of projectile motion.
In this scenario, the horizontal component of the airplane's velocity remains constant at 150 m/s, while the vertical component is not relevant to finding the horizontal distance. The time taken for the bomb to hit the target can be determined solely based on the vertical motion.
First, let's find the time taken for the bomb to reach the ground. We can use the equation:
h = (1/2) * g * t^2
Where h is the height (150m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.
Rearranging the equation and solving for t:
t^2 = (2h) / g
t^2 = (2 * 150) / 9.8
t^2 ≈ 30.61
t ≈ √30.61
t ≈ 5.53 seconds
So, it takes approximately 5.53 seconds for the bomb to hit the ground.
Next, we can find the horizontal distance traveled by the plane during this time using the equation:
d = v * t
Where d is the horizontal distance, v is the horizontal velocity of the plane (150 m/s), and t is the time taken (5.53 seconds).
Substituting the values:
d = 150 * 5.53
d ≈ 829.5 meters
Therefore, the horizontal distance of the plane from the target when the bomb was released is approximately 829.5 meters.
h = 0.5g*t^2 = 150 m.
4.9t^2 = 150
t^2 = 30.61
t = 5.53 s. = Fall time.
d = Xo * t = 150m/s * 5.53 s = 830 m.