A ball is thrown vertically upward with a velocity of \[60\,m/s\]. The maximum height to which ball rises is ?

Hint:The velocity, which is a function of time, is the rate at which an object's location changes with respect to a frame of reference. A specification of an object's speed and direction of motion is referred to as velocity.

Complete step by step answer:
The initial velocity is the object's velocity until acceleration changes it. The new velocity is the final velocity after accelerating for some time. In the question, it is given that initial velocity, \[u = 60\,m/s\]. The final velocity, \[v = 0\]. Because at the greatest height, where the ball can reach, the velocity is \[0\].

The acceleration due to gravity, \[g = 9.8\,m/{s^2}\]. We have to calculate the height of the ball to which it rises using these values. From the equation, \[{v^2} = {u^2} + 2aS\], we can find the maximum height, \[S\].After rearranging the formula, we get
\[S = \dfrac{{{v^2} - {u^2}}}{{2a}}\]
\[\Rightarrow S = \dfrac{{0 - {{\left( {60} \right)}^2}}}{{2 \times 9.8}} \\
\therefore S= 183.7\,m\]

So, \[183.7\,m\] is the maximum height to which the ball can rise.

Note:Remember the formula \[{v^2} = {u^2} + 2aS\] . We can use this formula to determine the initial velocity, final velocity, and height if one value is missing and the others are given in the question.An object must have a constant speed in a constant direction to have a constant velocity. The object's motion is restricted to a straight path when it has a constant direction. As a result, motion in a straight line at a constant speed is referred to as constant velocity.