Question

You throw a ball vertically upward so that it leaves the ground with velocity 6.48 m/s. (a) what is its velocity when it reaches its maximum altitude

Answer 1

the velocity is 0 m/s, as when it reaches the top, the object has no velocity for a small period of time as it then switched to a negative velocity when coming down

Answer 2

Answer:

The velocity of the ball when it reaches its maximum altitude is $v=0\frac{m}{s}$

Explanation:

This is a vertical shot, and we know that in this kind of motion, the particle (in this case the ball) starts with a given velocity, in this problem given by

$v_{0}=6.48\frac{m}{s}$

wich points upward in an imaginary vertical axis, and let's say that this is the positive direction. When that particle (ball) reaches the ground again due to the presence of gravity, it will have the same initial velocity, but with changed sign, as it is coming down, so

$v_{f}=-6.48\frac{m}{s}$

is the final velocity for the particle (ball) when it reaches the ground.

Now, in order for this to happen (the velocity has to change from a positive value to a negative value), there must be a point in the trajectory where it changes its direction.

This point (wich is where the ball stops going upwards and starts going downwards, and this happens in an instant) coincides with the point of maximum altitude.

So, the answer is that there is an instant when the ball changes from positive velocity to negative velocity, where its velocity is zero and that is the point of maximum altitude in its trajectory.