A ball is projected vertically upward with a speed of $50\;{\text{m/s}}$ . Find
a) The maximum height
b) The time to reach the maximum height
c) The speed at half the maximum height. Take $g = 10\;{\text{m/}}{{\text{s}}^{\text{2}}}$

Hint: The equations of motion are used for finding the unknown variables like velocity, time, and acceleration. For the projectile motions, the acceleration is the acceleration due to gravity and the displacement is the vertical height is reached. Hence using the equations of motion we can find the maximum height and time taken to reach the height.

Complete step by step answer:
Given the initial velocity of the ball is $u = 50\;{\text{m/s}}$

(a) Let’s consider the equation of motion
${v^2} - {u^2} = 2as$
Where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement.
Taking the final velocity is zero when it reaches the maximum height and the displacement as the maximum height, we can modify the equation as follows
$
  {0^2} - {u^2} = 2 \times - g \times {h_{\max }} \\
  {h_{\max }} = \dfrac{{{u^2}}}{{2g}} \\
$
Substituting the values in the above expression,
$
{h_{\max }} = \dfrac{{{{\left( {50\;{\text{m/s}}} \right)}^2}}}{{2 \times 10\;{\text{m/}}{{\text{s}}^{\text{2}}}}} \\
   = 125\;{\text{m}} \\
$
Thus the maximum height the ball can reach will be $125\;{\text{m}}$.

(b) We can apply the first equation of motion to find the time taken to reach the maximum height.
The equation is
$v = u + at$
Where, $t$ is the time taken, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration.
When the ball reaches maximum height the velocity becomes zero. Hence the above equation changes to,
$
  0 = u - g \times t \\
  t = \dfrac{u}{g} \\
$
Substitute the values in the above expression,
$
  t = \dfrac{{50\;{\text{m/s}}}}{{10\;{\text{m/}}{{\text{s}}^{\text{2}}}}} \\
   = 5\;{\text{s}} \\
$
Thus the ball takes $5\;{\text{s}}$ to reach the maximum height.

(c) We have to find the speed at half the maximum height. Let’s take the velocity at $\dfrac{{{h_{\max }}}}{2}$ height as $V'$ .
Applying these variables to the third equation of motion, we get
 $
  {{v'}^2} - {u^2} = 2 \times - g \times \dfrac{{{h_{\max }}}}{2} \\
  {{v'}^2} = 2 \times - g \times \dfrac{{{h_{\max }}}}{2} + {u^2} \\
$
Substituting the values in the above expression,
$
  {{v'}^2} = 2 \times - 10\;{\text{m/}}{{\text{s}}^{\text{2}}} \times \dfrac{{125\;{\text{m}}}}{2} + {\left( {50\;{\text{m/s}}} \right)^2} \\
   = 1250\;{{\text{m}}^2}{\text{/}}{{\text{s}}^{\text{2}}} \\
$
Taking the square root,
$
  v' = \sqrt {1250} \;{\text{m/s}} \\
  {\text{ = 35}}{\text{.35}}\;{\text{m/s}} \\
$

Thus the speed of the ball when it reaches the half of maximum height is ${\text{35}}{\text{.35}}\;{\text{m/s}}$.

Note:
While solving the equations of projectile motion, the acceleration due to gravity is taken to be negative. Since the displacement is vertically upwards which is against gravity. And at the maximum height, we can consider the final velocity as zero for easy calculation.