A dart is thrown horizontally toward the center point X of the target at 20 m/s. After

0.1 s it hits a point Y which lies exactly below
point X.
The distance XY is:
1. 1 m
2. 0.1 m
3. 0.5 m
4. 2 m
5. 0.05 m

the answer is 5) 0.05

To find the distance XY, we can use the formula:

XY = Vx * t

Where Vx is the horizontal velocity of the dart and t is the time taken to hit point Y.
Since the dart is thrown horizontally, the initial vertical velocity (Vy) is 0 m/s. The only force acting on the dart is gravity, which does not affect the horizontal motion.

Given:
Initial velocity (V) = 20 m/s
Time (t) = 0.1 s

We can find the horizontal velocity (Vx) using the formula:

Vx = V * cos(theta)

Since the dart is thrown horizontally, the angle of projection (theta) is 0°, and the cosine of 0° is 1.

Vx = V * cos(0) = V

Vx = 20 m/s

Now, we can substitute the values of Vx and t into the formula:

XY = Vx * t = 20 m/s * 0.1 s = 2 m

Therefore, the distance XY is 2 m. Hence, the correct option is 4. 2 m.

To determine the distance XY, we need to understand the motion of the dart.

The dart is thrown horizontally, which means it only has an initial velocity in the horizontal direction. Therefore, the vertical velocity of the dart is 0 m/s.

We can use the equation of motion in the vertical direction to determine the time it takes for the dart to hit point Y:

Y = Y0 + V0yt - 0.5gt^2

Since the dart is thrown at point X, which is the same height as Y (i.e., Y0 = 0), and the vertical velocity V0y = 0 m/s, the equation simplifies to:

Y = -0.5gt^2

Now, we can substitute the given time t = 0.1 s into the equation to find the vertical distance:

Y = -0.5 * 9.8 m/s^2 * (0.1 s)^2

Y = -0.5 * 9.8 m/s^2 * 0.01 s^2

Y = -0.049 m

The negative sign indicates that the dart has fallen below the starting point X.

So, the distance XY is equal to the absolute value of Y, which is 0.049 m, or approximately 0.05 m.

Therefore, the correct answer is 5. 0.05 m.

You try. I showed you how.