A ball with a mass of 3 kg is thrown upwards with a velocity of 5 m/s, what is the maximum height the ball will reach

1.27 m

Explanation:

Relating kinetic energy and potential energy by following the law of conservation of linear energy then we can say where m is mass, g is acceleration due to gravity, h is height which is unknlwn in this question, v is the velocity. Making h the subject of the above formula we get that and substituting g with 9.81 m/s2, v with 5 m/s we can deduce that [tex]h=\frac {0.5\times 5^{2}}{9.81}=1.2742099898063[\approx 1.27 m/tex]