A helicopter is ascending vertically with a speed of 6.50 m/s. At a height of 55 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: The package's initial speed equals the helicopter's.]
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To find the time it takes for the package to reach the ground, we can use the equation of motion for vertical motion:
s = ut + (1/2)gt^2
Where:
s = height (distance) traveled
u = initial velocity
t = time
g = acceleration due to gravity
In this case, the helicopter is ascending vertically with a speed of 6.50 m/s, which means the initial velocity of the package is also 6.50 m/s. The height traveled by the package is 55 m.
We know that the acceleration due to gravity, g, is approximately 9.8 m/s^2.
Rearranging the equation to solve for time:
55 = 6.50t + (1/2)(9.8)t^2
Simplifying the equation, we get a quadratic equation:
4.9t^2 + 6.50t - 55 = 0
To solve the quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this quadratic equation:
a = 4.9, b = 6.50, and c = -55
Substituting these values into the quadratic formula:
t = (-6.50 ± √(6.50^2 - 4 * 4.9 * -55)) / (2 * 4.9)
Solving this equation, we get two solutions, but we discard the negative value since time cannot be negative:
t = (-6.50 + √(6.50^2 - 4 * 4.9 * -55)) / (2 * 4.9)
Calculating this value using a calculator, we find that:
t = 3.67 seconds (rounded to two decimal places)
Therefore, it takes approximately 3.67 seconds for the package to reach the ground.