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:

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