A military helicopter on a training mission is flying horizontally at a speed of 70.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 320 m. You can ignore air resistance.
(a) How much time is required for the bomb to reach the earth?
(b) How far does it travel horizontally while falling?
(c) Find the horizontal and vertical components of its velocity just before it strikes the earth.
(horizontal component)
(vertical component)
(e) If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?
height above ground
a) how long does it take to fall 320m
320=1/2 g t^2
b) distance=horizontalvelocity*timeinair
c)horizontal, still the same. vertical= g*t
e) The helicopter has the same horizontal velocity as the bomb.
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To solve this problem, we can use the equations of motion and the principles of projectile motion. Let's break it down step by step:
(a) To find the time required for the bomb to reach the earth, we can use the equation of motion for vertical free fall:
s = ut + (1/2)gt^2
where:
s = vertical distance traveled (320 m)
u = initial vertical velocity (0 m/s since the bomb starts at rest)
t = time taken
g = acceleration due to gravity (-9.8 m/s^2)
Plugging in the given values, the equation becomes:
320 = 0*t + (1/2)*(-9.8)*t^2
Simplifying:
320 = -4.9t^2
Divide both sides by -4.9:
t^2 = -320/-4.9
t^2 = 65.3
Taking the square root of both sides, we get:
t ≈ 8.075 seconds
Therefore, it takes approximately 8.075 seconds for the bomb to reach the earth.
(b) To find the horizontal distance traveled while falling, we can use the equation of motion for horizontal motion:
s = ut
where:
s = horizontal distance traveled
u = horizontal velocity (since the helicopter is flying horizontally at 70.0 m/s, the bomb inherits this velocity)
t = time (8.075 seconds, as calculated above)
Plugging in the values:
s = 70.0 * 8.075
s ≈ 565.25 meters
Therefore, the bomb travels approximately 565.25 meters horizontally while falling.
(c) Just before the bomb strikes the earth, the vertical component of its velocity will be equal to the final vertical velocity after falling for 8.075 seconds, given by:
v = u + gt
where:
v = final vertical velocity
u = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (8.075 seconds)
Plugging in the values:
v = 0 + (-9.8) * 8.075
v ≈ -79.19 m/s (Negative sign indicates downward direction)
Therefore, the vertical component of the velocity just before striking the earth is approximately -79.19 m/s.
The horizontal component of the velocity remains constant throughout the motion and is equal to the horizontal velocity of the helicopter, which is 70.0 m/s.
(e) Since the velocity of the helicopter remains constant, it means that the helicopter maintains its position during the bomb's fall. Therefore, the helicopter is still at the same horizontal position when the bomb hits the ground.
However, to find the height of the helicopter above the ground at that time, we can use the equation of motion for vertical motion:
s = ut + (1/2)gt^2
where:
s = vertical distance traveled by the helicopter (height above the ground)
u = initial vertical velocity (0 m/s, assuming the helicopter's altitude remains constant)
t = time taken (8.075 seconds, as calculated above)
g = acceleration due to gravity (-9.8 m/s^2)
Plugging in the values:
s = 0 * 8.075 + (1/2) * (-9.8) * (8.075)^2
s ≈ -251.76 meters (Negative sign indicates the height below the starting altitude)
Therefore, the helicopter is approximately 251.76 meters below its initial altitude when the bomb hits the ground.