Package is released from a helicopter that is rising upward.?
At the instant of its release, the object is located 50 m above the ground. Just before the package hits the ground, it is traveling straight down at a speed of 33.6 m/s. Assume air resistance is negligible. Answer from the viewpoint of a person standing at rest on the ground.
How fast was the package moving at the instant of its release from the helicopter?
How much time passes from the instant of release of the package until the instant just before it hits the ground?
At the instant of its release, the object is located 50 m above the ground. Just before the package hits the ground, it is traveling straight down at a speed of 33.6 m/s. Assume air resistance is negligible. Answer from the viewpoint of a person standing at rest on the ground.
How fast was the package moving at the instant of its release from the helicopter?
How much time passes from the instant of release of the package until the instant just before it hits the ground?
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Package is released from a helicopter that is rising upward.?
At the instant of its release, the object is located 50 m above the ground. Just before the package hits the ground, it is traveling straight down at a speed of 33.6 m/s. Assume air resistance is negligible. Answer from the viewpoint of a person standing at rest on the ground.
Let upward motion be positive. The acceleration of the package is -9.8 m/s^2. The initial vertical position = 50 m. The final vertical position = 0 m. The final vertical velocity = -33.6 m/s.
Final velocity^2 = Initial velocity^2 + (2 * acceleration * displacement)
Displacement = Final position – Initial position = 0 – 50 = -50 m
(-33.6)^2 = Initial velocity^2 + (2 * -9.8 * -50)
Initial velocity^2 = (-33.6)^2 – (2 * -9.8 * -50) = 1128.96 – 980 = 148.96
Initial velocity = √148.96 = ± 12.2 m/s
Since the package was moving upward, the initial velocity = +12.2 m/s
Final velocity = Initial velocity + acceleration * time
Time = Final velocity – Initial velocity ÷acceleration
Time = (-33.6 - +12.2) ÷ -9.8 = 4.67 seconds
Let’s check the answer!
As the package rises its speed decreases from 12.2 m/s to 0 m/s at the rate of 9.8 m/s each second.
Time up = 12.2 ÷ 9.8 = 1.245 seconds
Distance up = 12.2 * 1.245 – (½ * 9.8 * 1.245^2) = 7.594 meters
Now the package is 57.594 meters above the ground.
At the instant of its release, the object is located 50 m above the ground. Just before the package hits the ground, it is traveling straight down at a speed of 33.6 m/s. Assume air resistance is negligible. Answer from the viewpoint of a person standing at rest on the ground.
Let upward motion be positive. The acceleration of the package is -9.8 m/s^2. The initial vertical position = 50 m. The final vertical position = 0 m. The final vertical velocity = -33.6 m/s.
Final velocity^2 = Initial velocity^2 + (2 * acceleration * displacement)
Displacement = Final position – Initial position = 0 – 50 = -50 m
(-33.6)^2 = Initial velocity^2 + (2 * -9.8 * -50)
Initial velocity^2 = (-33.6)^2 – (2 * -9.8 * -50) = 1128.96 – 980 = 148.96
Initial velocity = √148.96 = ± 12.2 m/s
Since the package was moving upward, the initial velocity = +12.2 m/s
Final velocity = Initial velocity + acceleration * time
Time = Final velocity – Initial velocity ÷acceleration
Time = (-33.6 - +12.2) ÷ -9.8 = 4.67 seconds
Let’s check the answer!
As the package rises its speed decreases from 12.2 m/s to 0 m/s at the rate of 9.8 m/s each second.
Time up = 12.2 ÷ 9.8 = 1.245 seconds
Distance up = 12.2 * 1.245 – (½ * 9.8 * 1.245^2) = 7.594 meters
Now the package is 57.594 meters above the ground.
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