A cargo plane is moving with a horizontal velocity of
vx = +230 m/s
at a height of
y = 840 m
above level ground as shown in the figure below when it releases a package. Ignoring air resistance, how much time will it take the package to reach the ground? (Express your answer to the nearest tenth of a second.)
s
h = Vy*t + 0.5g*t^2 = 840 m.
0*t + 4.9t^2 = 840
Solve for t.
To solve this problem, we can use the equations of motion to find the time it takes for the package to reach the ground. Since the package is released from a certain height above the ground, we can assume that its initial vertical velocity is zero.
First, let's find the time it takes for the package to reach the ground using the equation of motion for vertical displacement:
y = y0 + v0y * t + (1/2) * a * t^2
where:
y = vertical displacement (840 m)
y0 = initial vertical position (0 m)
v0y = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)
Simplifying the equation, we get:
840 = 0 + 0 * t + (1/2) * (-9.8) * t^2
840 = -4.9 * t^2
Dividing both sides by -4.9:
-171.4 = t^2
Taking the square root of both sides:
t = √171.4
Using a calculator, we find:
t ≈ 13.1 seconds
Therefore, it will take approximately 13.1 seconds for the package to reach the ground.