A stone is thrown vertically upwards with an initial velocity of 40m/s

1.	A stone is thrown vertically upwards with an initial velocity of 40m/s.Find the maximum height reached by the stone. What is the netdisplacement and total distance covered by the stone?VERY URGENT PLS ANSWER GR 9 QUESTION
Answer» Answer: 81.63 meters Explanation: We will USE the following equation ${v}^{2} - {u}^{2} = 2as$ $initial \: <klux>VELOCITY</klux> \: = 40m {sec}^{ - 1}$ At MAX height,final velocity is ZERO ${0}^{2} - {40}^{2} = 2 \times ( - 9.8) \times s$ (Negative Acceleration) Solving, $s = \frac{1600}{19.6}$ $s = 81.63m$ Total distance is 163.26 m Total displacement is zero