Answer to Question #278088 in Physics for pogs

Question

A ball is thrown upward with an initial velocity of 10m/s. how high does the ball rise? and what is the total time flight

Answer 1

Let's first find the time that the ball takes to reach maximum height:

$$v=v_0-gt,$$ $$0=v_0-gt$$ $$t=\dfrac{v_0}{g}=\dfrac{10\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=1.02\ s.$$

Then, we can find the maximum height reached by the ball from the kinematic equation:

$$y_{max}=v_0t-\dfrac{1}{2}gt^2,$$ $$y_{max}=10\ \dfrac{m}{s}\times1.02\ s-\dfrac{1}{2}\times9.8\ \dfrac{m}{s^2}\times(1.02\ s)^2=5.1\ m.$$

Finally, we can find the total time flight as follows:

$$t_{flight}=2t=2\times1.02\ s=2.04\ s.$$