91Ó°ÊÓ

• The starting velocity of the ball in horizontal direction is given as\(16\,m/s\).
• The initial velocity of the ball in vertical direction is given as \(12\,m/s\).

The expression for the final velocity from Newton’s third law of motion is given by,

\({\left( {{v_f}} \right)^2} = {\left( {{v_i}} \right)^2} + 2ay\)

Here\(a\)is the acceleration,\({v_f}\)is the final velocity,\({v_i}\)is the initial velocity,\(t\)is required time,\(y\)is the maximum height attained by the ball.

The expression for the maximum height attained by the ball is obtained from the third law of motion:

\({\left( {{v_f}} \right)^2} = {\left( {{v_i}} \right)^2} + 2ay\)

Solve for\({y_{\max }}\)

\({y_{\max }} = \frac{{{{\left( {{v_f}} \right)}^2} - {{\left( {{v_i}} \right)}^2}}}{{2a}}\)

Substitute \(0\,m/s\) for \({v_f}\), \(12\,m/s\) for \({v_i}\) and \( - 9.8\,m/{s^2}\) for \(a\) into the above equation,

\(\begin{aligned}{}{y_{\max }} &= \frac{{{{\left( {0\;m/s} \right)}^2} - {{\left( {12\;m/s} \right)}^2}}}{{2\left( { - 9.81\;m/{s^2}} \right)}}\\ &= 7.34\,m\end{aligned}\)

Therefore the maximum height attained by the ball is \(7.34\;m\).