Answer to Question #158664 in Classical Mechanics for Lorna

Question

Find the maximum height reached by an object projected vertically upwards with an initial velocity of 40 meters per second and find the time taken for that

Answer 1

Let's first find the time that the object takes to reach the maximum height from the kinematic equation:

$$v=v_0-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}=\dfrac{40\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=4.1\ s.$$

Finally, we can find the maximum height that the object can reach from the kinematic equation:

$$y=v_0t-\dfrac{1}{2}gt^2,$$ $$y=40\ \dfrac{m}{s}\cdot4.1\ s-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(4.1\ s)^2=81.6\ m.$$