A helicopter ascending at a rate of 12 m/s . At a height of 80 m above the ground a package is dropped . How long does the package take to reach the ground ?
Solution
Step 1: use second equation of motion.
Step 2: use quadratic equation.
just the normal one. A projectile shot from a height h with initial velocity v has height
h(t) = h + vt - 4.9t^2
Now just solve for t when h=0.
I get 5.45 seconds
To find the time it takes for the package to reach the ground, we can use the equation of motion for free fall:
h = (1/2) * g * t^2
Where:
h = height (80 m in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since the package is dropped from rest, its initial velocity is 0 m/s. Therefore, we can ignore the term involving initial velocity in the equation:
h = (1/2) * g * t^2
Rearranging the equation, we get:
t^2 = (2 * h) / g
Now, substituting the given values:
t^2 = (2 * 80) / 9.8
t^2 = 16.3265
To find t, we take the square root of both sides:
t ≈ √16.3265
t ≈ 4.04 seconds
Therefore, the package takes approximately 4.04 seconds to reach the ground.