Thank you for visiting A small mailbag is released from a helicopter that is descending steadily at 1 70 m s a After 5 00 s what is the. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
After 5.00 s, the speed of the mailbag is 47.3 m/s.
After 5.00 s, the mailbag is 114.0 m below the helicopter.
After 5.00 s, the speed of the mailbag would be 50.7 m/s.
After 5.00 s, the mailbag would be 131.0 m below the helicopter if the helicopter is rising steadily at 1.70 m/s.
(a) After 5.00 s, the speed of the mailbag can be determined using the principle of relative motion. Since the mailbag is released from a descending helicopter, its initial velocity is the same as the helicopter's descending velocity.
Given:
Descending velocity of the helicopter = -1.70 m/s (negative sign indicates downward direction)
Time = 5.00 s
To find the speed of the mailbag, we need to consider the relative motion between the mailbag and the descending helicopter. The mailbag's speed is the vector sum of its initial velocity (helicopter's descending velocity) and the acceleration it experiences due to gravity.
Using the equation:
Speed = Initial velocity + (Acceleration due to gravity × Time)
Speed = -1.70 m/s + (9.8 m/s² × 5.00 s) [Acceleration due to gravity is positive since it opposes the downward motion]
Calculations:
Speed = -1.70 m/s + (9.8 m/s² × 5.00 s)
Speed = -1.70 m/s + 49.0 m/s
Speed = 47.3 m/s
Therefore, after 5.00 s, the speed of the mailbag is 47.3 m/s.
(b) To determine how far below the helicopter the mailbag is after 5.00 s, we can use the equation of motion:
Distance = Initial velocity × Time + (1/2) × Acceleration × Time²
Given:
Descending velocity of the helicopter = -1.70 m/s
Time = 5.00 s
Using the equation:
Distance = (-1.70 m/s × 5.00 s) + (0.5 × 9.8 m/s² × (5.00 s)²)
Calculations:
Distance = (-1.70 m/s × 5.00 s) + (0.5 × 9.8 m/s² × (5.00 s)²)
Distance = -8.50 m + 122.5 m
Distance = 114.0 m
Therefore, after 5.00 s, the mailbag is 114.0 m below the helicopter.
(c) If the helicopter is rising steadily at 1.70 m/s, the calculations will be different.
(a) After 5.00 s, the speed of the mailbag would be the vector sum of its initial velocity (helicopter's rising velocity) and the acceleration due to gravity.
Speed = 1.70 m/s + (9.8 m/s² × 5.00 s) [Acceleration due to gravity is positive since it opposes the upward motion]
Calculations:
Speed = 1.70 m/s + (9.8 m/s² × 5.00 s)
Speed = 1.70 m/s + 49.0 m/s
Speed = 50.7 m/s
Therefore, after 5.00 s, the speed of the mailbag would be 50.7 m/s.
(b) To determine how far below the helicopter the mailbag is after 5.00 s, we use the equation of motion again:
Distance = (Initial velocity × Time) + (0.5 × Acceleration × Time²)
Given:
Rising velocity of the helicopter = 1.70 m/s
Time = 5.00 s
Using the equation:
Distance = (1.70 m/s × 5.00 s) + (0.5 × 9.8 m/s² × (5.00 s)²)
Calculations:
Distance = (1.70 m/s × 5.00 s) + (0.5 ×
9.8 m/s² × (5.00 s)²)
Distance = 8.50 m + 122.5 m
Distance = 131.0 m
Therefore, after 5.00 s, the mailbag would be 131.0 m below the helicopter if the helicopter is rising steadily at 1.70 m/s.
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