A package is dropped from a helicopter moving upward at 15 m/s. If it takes 12 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?A package is dropped from a helicopter moving upward at 15 m/s. If it takes 12 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?318 m636 m530 m424 m

Respuesta :

Answer:

option C

Explanation:

given,

helicopter moving at the speed of  = 15 m/s

time before the package hit the ground = 12 s

height above the ground = ?

using the equation

[tex]h_f = h_i + u t + \dfrac{1}{2}gt^2[/tex]

[tex]0 = h_i + 15 \times 12 - \dfrac{1}{2}\times 9.8 \times 12^2[/tex]

[tex]0 = h_i + 180 - \dfrac{1}{2}\times 9.8 \times 12^2[/tex]

[tex] h_i = 526.32\ m[/tex]

hence, the height above the ground will be approximately equal to 530 m

correct answer is option C

The kinematic equation of motion gives the interrelationships of the variables of motion

The correct option for the distance the package above the ground when it was released, is the third option;

530 m

The reason the above selection is correct is as follows:

The known parameters are;

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

Method:

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

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