A helicopter is ascending vertically with a speed of 5.20 m/s. At a height of 120 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: v_0 for the package equals the speed of the helicopter.]
h=vi*t-1/2 g t^2 solve for t.
To find the time it takes for the package to reach the ground, we can use the equation of motion:
s = v₀t + (1/2)at²
Where:
s = height traveled (in this case, 120 m downwards since the package is dropped)
v₀ = initial velocity (in this case, the speed of the helicopter, which is 5.20 m/s)
t = time taken
a = acceleration (in this case, acceleration due to gravity, approximately 9.8 m/s²)
Since the package is dropped, its initial velocity is the same as the speed of the helicopter, 5.20 m/s, but in the opposite direction (downwards), so we can take v₀ as -5.20 m/s.
Plugging in the values into the equation:
-120 = -5.20t + (1/2)(-9.8)t²
Simplifying the equation:
-120 = -5.20t - 4.9t²
To find t, we can rearrange the equation to form a quadratic equation:
4.9t² + 5.20t - 120 = 0
Now we can solve this quadratic equation using various methods, such as factoring, completing the square, or using the quadratic formula. Let's solve it using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values:
t = (-(5.20) ± √((5.20)² - 4(4.9)(-120))) / (2(4.9))
Calculating the value inside the square root:
t = (-5.20 ± √(27.04 + 2352)) / 9.8
t = (-5.20 ± √2379.04) / 9.8
t ≈ (-5.20 ± 48.78) / 9.8
Solving for the two possible values of t:
t₁ ≈ (43.58) / 9.8 ≈ 4.45 s
t₂ ≈ (-54.98) / 9.8 ≈ -5.61 s
Since time can't be negative, we discard the negative value for t. Therefore, the time it takes for the package to reach the ground is approximately 4.45 seconds.