A cargo plane is flying horizontally at an
altitude of 11.2 km with a constant speed of
1260 km/hr when a large crate slides out of
the rear of the plane.
How long does it take the crate to hit
the ground? The acceleration of gravity is
9.81 m/s
2
.
Answer in units of s, please.
h = 11.2km = 11,200 m.
h = 0.5g*t^2 = 11,200
4.9*t^2 = 11,200
t^2 = 2286
Tf = 47.8 s. = Fall time.
To find the time it takes for the crate to hit the ground, we can use the equation of motion for vertical motion.
The initial vertical velocity of the crate when it slides out of the plane is zero because the crate is initially at rest relative to the plane. The final vertical displacement is the altitude of the plane, -11.2 km. The acceleration acting on the crate is the acceleration due to gravity, which is -9.81 m/s^2 (negative because it acts in the opposite direction to the positive vertical axis convention).
The equation of motion for vertical motion is:
s = ut + (1/2)at^2
where:
- s is the vertical displacement
- u is the initial vertical velocity (which is zero in this case)
- a is the acceleration due to gravity
- t is the time
Plugging in the known values:
- s = -11.2 km = -11.2 × 10^3 m (converting km to m)
- u = 0 m/s
- a = -9.81 m/s^2
We want to solve for t. Rearranging the equation, we get:
t = sqrt((2s)/a)
Substituting in the values:
t = sqrt((2 * -11.2 × 10^3 m) / (-9.81 m/s^2))
Calculating this expression, we find that t ≈ 14.80 seconds.
Therefore, it takes approximately 14.80 seconds for the crate to hit the ground.