Answer:
We can start by determining the time it takes for the ball to fall from the top of the building to the ground. We can use the equation:
y = 0.5gt^2
where y is the vertical distance traveled by the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. The initial vertical velocity of the ball is 0 m/s, since it is dropped from rest. The vertical distance traveled by the ball is the height of the building, which is 50.0 m. Substituting these values, we get:
50.0 m = 0.5(9.8 m/s^2)t^2
t = √(50.0 m / (0.5 × 9.8 m/s^2))
t = 3.19 s (to two decimal places)
So, it takes approximately 3.19 seconds for the ball to fall from the top of the building to the ground.
Next, we can determine the horizontal distance traveled by Person A during this time. The horizontal distance is given by:
d = vt
where d is the distance traveled, v is the velocity, and t is the time. Substituting the given values, we get:
d = (0.47 m/s)(3.19 s)
d = 1.50 m (to two decimal places)
So, Person A moves approximately 1.50 meters horizontally during the time it takes for the ball to fall from the top of the building to the ground.
Since the ball lands 1.0 meter in front of the entrance, and Person A is 3.0 meters away from the entrance, the ball will not land on Person A. Therefore, Person A is safe from the falling ball.
Step-by-step explanation: