An object was launched with a velocity of 20 ms−1 at an angle of 45° to the vertical. At the

top of its trajectory the object broke into two equal pieces. One piece fell vertically
downwards. Where would the other piece fall? (Take g = 10 ms−2) (

To solve this problem, we need to break it down into different components and use some basic principles of projectile motion.

1. Initial velocity: The object was launched with a velocity of 20 m/s at an angle of 45° to the vertical. We can split this velocity into horizontal and vertical components.

The horizontal component of the velocity can be found using cosine:
Vx = V * cos(θ) = 20 m/s * cos(45°) = 20 m/s * √2 / 2 = 10√2 m/s

The vertical component of velocity can be found using sine:
Vy = V * sin(θ) = 20 m/s * sin(45°) = 20 m/s * √2 / 2 = 10√2 m/s

2. Time to reach the top: At the top of the trajectory, the vertical component of velocity becomes zero. We can use the formula for vertical motion to find the time it takes to reach the top:
Vy = V0y + gt
0 = 10√2 - 10*t
t = 1/√2 seconds

3. Height at the top: To find the height at the top, we can use the formula for vertical motion:
y = V0y*t - (1/2)gt^2
y = 10√2 * (1/√2) - (1/2)*10*(1/√2)^2
y = 10 - 5/2
y = 15/2 = 7.5 meters

4. Horizontal distance: Since the horizontal velocity remains constant throughout the motion, the horizontal distance traveled can be found by multiplying the horizontal velocity by the time:
x = Vx * t = 10√2 * (1/√2) = 10 meters.

5. Falling vertically: One piece falls vertically downwards. Since gravity exerts a downward acceleration of 10 m/s^2, this piece will fall straight downwards without any horizontal motion.

6. Where does the other piece fall? Since the object broke into two equal pieces and one falls vertically downwards, the other piece will also fall at the same horizontal distance from the launch point. Therefore, the other piece will fall 10 meters away horizontally from the launch point.

To summarize, the other piece will fall 10 meters away from the launch point horizontally and at the same height as the top of the trajectory, which is 7.5 meters above the launch point.