Question

An object is dropped from a helicopter which is moving horizontally at a constant velocity of 45m/s^1 180m above the ground. What is the time taken for the object to reach the ground?

Answer 1

The time taken for the object to reach the ground does not depend on the initial horizontal speed and equals:

$t=\sqrt{\frac{2h}{g} } =\sqrt{\frac{2\cdot 180}{9.8} } =6.06\space s$