Question

An object is dropped from a helicopter which is moving horizontally at a constant velocity of 45 m/s 180m above the ground. Find the time taken for the object to reach the ground.

Answer 1

Answer on Question #49944 – Physics - Mechanics | Kinematics | Dynamics

An object is dropped from a helicopter which is moving horizontally at a constant velocity of 45 m/s 180m above the ground. Find the time taken for the object to reach the ground.

Solution:

$h = 180\,m$ – height of the helicopter;

$V = 45\,\frac{m}{s}$ velocity of the helicopter;

Equation of motion for the object along $Y - axis$:

$$y: h = \frac{g t^2}{2}$$ $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 180\,m}{9.8\, \frac{m}{s^2}}} = 6\,s$$

Answer: time taken for the object to reach the ground is equal to 6 s.

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