A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s when it accidentally drops a bomb (fortunately, not armed) at an elevation of 300 m. You can ignore air resistance.
How much time is required for the bomb to reach the earth?
7.82s
How far does it travel horizontally while falling?
469m
Find the horizontal component of the bomb's velocity just before it strikes the earth.
60m
What is the vertical component of the bomb's velocity just before it strikes earth?
If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?
To find the vertical component of the bomb's velocity just before it strikes the earth, we can use the equation:
v² = u² + 2as
Where:
v = final velocity (vertical component of the bomb's velocity just before it strikes)
u = initial velocity (initial vertical component of the bomb's velocity)
a = acceleration due to gravity (approximately -9.8 m/s², negative because it acts downward)
s = distance traveled vertically (300 m)
Since the bomb was dropped with no initial vertical velocity (u = 0):
v² = 0² + 2(-9.8)(300)
v² = 2(-9.8)(300)
v² = -5880
v = √(-5880)
v ≈ -76.73 m/s
The negative sign indicates that the velocity is downward.
Since the velocity is a vector quantity, both the horizontal and vertical components contribute to the overall velocity. Therefore, the horizontal component of the bomb's velocity just before it strikes the earth remains the same as the initial horizontal velocity of the helicopter, which is 60.0 m/s.
Now, let's determine where the helicopter is when the bomb hits the ground. Since the helicopter is flying at a constant speed horizontally, the horizontal distance covered by the helicopter is the same as the horizontal distance traveled by the bomb.
To find this distance, we can use the formula:
s = ut
Where:
s = distance traveled horizontally (to be found)
u = initial horizontal velocity (60.0 m/s)
t = time taken for the bomb to reach the earth (to be found)
Rearranging the formula, we get:
t = s / u
Substituting the values known, we have:
t = 469 m / 60.0 m/s
t ≈ 7.82 s
Therefore, when the bomb hits the ground after 7.82 seconds, the helicopter would have covered a horizontal distance of 469 meters.