From a helicopter descending uniformly at a speed of 2 m / s, a shot was fired vertically upward.

From a helicopter descending uniformly at a speed of 2 m / s, a shot was fired vertically upward. The bullet flew out at a speed of 200 m / s relative to the ground. How long will it take for the bullet to catch up with the helicopter?

Let us direct the y-axis, along which we will plot the coordinates of the bullet and the helicopter, upward:

The initial bullet velocity vp in this case will be positive, and the gravitational acceleration g will be negative.

Bullet motion law (dependence on time t):

yп = vпt – gt ^ 2/2.

The helicopter speed vw in this frame of reference will be negative:

yв = -vвt.

At the moment when the bullet is equal to the helicopter, their y coordinates should be equal:

yв = yп;

-vвt = vпt – gt ^ 2/2.

It is clear that the coordinates are equal at the initial moment t = 0. Let us cancel the equality by t and find another solution:

-vv = vp – gt / 2;

-2vv = 2vp – gt;

-4 m / s = 400 m / s – 9.8 m / s2 * t;

9.8 m / s2 * t = 404 m / s;

9.8t = 404 s;

t = (404 / 9.8) c = 41.22 s.

Answer. The bullet and the helicopter will level out in 41.22 seconds.