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?
vi=8m/s
g= -9.8m/s^2
hf=hi+Vi*t-1/2 9.8t^2 solve for t to hit ground
how fast?
Vf=Vi-9.8 t
To find the time it takes for the package to hit the ground, we can use the equation of motion:
h = u*t + (1/2) * g * t^2
Where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
In this case, the helicopter is ascending with a constant velocity, so the initial velocity of the package is also 8 m/s upwards. The acceleration due to gravity is -9.8 m/s^2 because it acts downwards.
Substituting the values into the equation, we have:
-90 = 8t - (1/2) * 9.8 * t^2
Simplifying the equation gives us a quadratic equation:
-4.9t^2 + 8t - 90 = 0
To solve this equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Where a, b, and c are the coefficients of the quadratic equation. In this case, a = -4.9, b = 8, and c = -90.
Substituting these values into the quadratic formula, we have:
t = (-8 ± √(8^2 - 4(-4.9)(-90))) / 2(-4.9)
Simplifying further gives us two possible solutions for time:
t = (-8 ± √(64 - 1764)) / -9.8
t = (-8 ± √(-1700)) / -9.8
Since we cannot take the square root of a negative number, it means that the package will never hit the ground.