The distance the package above the ground when it was released, s ≈ 530 meters
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;
It is given that:
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
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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