When a ball is thrown vertically into the air, its velocity will become zero at the highest point of its trajectory. At this point, the ball reaches its maximum height and starts descending.
To determine the height at which the velocity becomes zero, we can use the equations of motion. The equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (s) is:
vf^2 = vi^2 + 2as
In this case, we're interested in the point where the velocity (vf) becomes zero, so we can rearrange the equation to solve for the displacement (s):
0 = vi^2 + 2as
Solving for s:
s = -vi^2 / (2a)
Since we're considering a ball thrown vertically into the air, the initial velocity (vi) is positive, and the acceleration (a) due to gravity is also positive but acts in the opposite direction to the initial velocity.
Therefore, we can substitute the values into the equation as follows:
s = -(vi^2) / (2 * (-g))
Where g is the acceleration due to gravity, approximately 9.8 m/s^2 on Earth.
This equation simplifies to:
s = (vi^2) / (2g)
So, the height at which the velocity becomes zero is given by the displacement (s) and can be calculated using the equation above.
Please note that the value of vi depends on the initial velocity with which the ball was thrown vertically into the air.