A helicopter is ascending (moving upwards) at a constant velocity of 8 m/s. At a height of 90m, a passenger opens a window and drops a package to the ground below. How long does it take the package to hit the ground and how fast is it moving just before impact?
See 9-5-11, 9:58pm post for solution.
To determine the time it takes for the package to hit the ground, we can use the formula for the time of flight for an object in freefall motion:
Time = √(2h/g)
Where h is the initial height of the package and g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the given values, we have:
Time = √(2 * 90 m / 9.8 m/s²)
Time = √(180 / 9.8)
Time ≈ √18.367
Time ≈ 4.29 seconds (rounded to two decimal places)
Therefore, it takes approximately 4.29 seconds for the package to hit the ground.
To find the speed of the package just before impact, we can use the equation for the final velocity of an object in freefall motion:
Final Velocity = Initial Velocity + (Acceleration * Time)
Since the package is dropped from rest, the initial velocity is 0 m/s. The acceleration due to gravity remains the same, 9.8 m/s². And the time of flight is approximately 4.29 seconds.
Final Velocity = 0 m/s + (9.8 m/s² * 4.29 s)
Final Velocity ≈ 42.04 m/s (rounded to two decimal places)
Therefore, the package is moving at a speed of approximately 42.04 m/s just before impact.
To solve this problem, we need to consider the motion of the helicopter and the package separately.
First, let's calculate the time it takes for the package to hit the ground. We know that the package is dropped from a height of 90m. We can use the equation for gravitational free fall motion:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. We can rearrange this equation to solve for t:
t = sqrt((2 * h) / g)
Plugging in the values, we get:
t = sqrt((2 * 90) / 9.8) ≈ 4.25 seconds
So it takes approximately 4.25 seconds for the package to hit the ground.
Now let's calculate the speed of the package just before impact. Since the package is dropped, it will fall freely under the influence of gravity. The speed of the package just before impact will be equal to the speed it acquired during this free fall. We can use the equation:
v = g * t
where v is the speed and t is the time. Plugging in the values, we get:
v = 9.8 * 4.25 ≈ 41.65 m/s
So the package is moving at approximately 41.65 m/s just before impact.