To find the velocity of the ball when it reaches Paul, we can use the principles of projectile motion. Since Peter throws the ball horizontally, the vertical component of its velocity will be zero. We only need to consider the horizontal component.
Let's assume that the ball takes time 't' to reach Paul after being thrown by Peter. During this time, it will cover the horizontal distance between Peter's window and Paul, which is 9 meters.
The horizontal distance traveled by an object can be calculated using the formula: distance = velocity × time.
Since the vertical component of velocity is zero, we only need to consider the horizontal distance. Therefore, we can write:
9 meters = velocity × t
We know that the vertical distance between Peter's window and the ground is 3 meters, which means the ball will fall from a height of 3 meters during the time 't'. We can use the equation for free fall motion to find the time 't' in terms of the acceleration due to gravity 'g':
3 meters = 0.5 × g × t²
Simplifying the equation:
0.5 × g × t² = 3
t² = 6 / g
Now, we can substitute the value of 't²' into the equation we obtained for the horizontal distance:
9 meters = velocity × (6 / g)^(1/2)
To solve for the velocity 'velocity', we need to know the value of the acceleration due to gravity 'g'. On Earth, 'g' is approximately 9.8 m/s². Plugging in this value:
9 meters = velocity × (6 / 9.8)^(1/2)
Simplifying further:
9 meters = velocity × (0.6122)
Dividing both sides by 0.6122:
velocity = 9 meters / 0.6122
velocity ≈ 14.7 meters per second (rounded to one decimal place)
Therefore, the velocity of the ball until it reaches Paul is approximately 14.7 meters per second.