Question

What maximum height attained by the object that is projected vertically upward with a
speed of 60 m/s? How long will it take for the object to reach the maximum height?

Answer 1

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

$$v=v_0-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}.$$

Then, we can substitute $t$ into another kinematic equation and find $y_{max}$:

$$y_{max}=v_0t-\dfrac{1}{2}gt^2,$$ $$y_{max}=v_0\dfrac{v_0}{g}-\dfrac{1}{2}g(\dfrac{v_0}{g})^2,$$ $$y_{max}=\dfrac{v_0^2}{2g}=\dfrac{(60\ \dfrac{m}{s})^2}{2\cdot 9.8\ \dfrac{m}{s^2}}=183.7\ m.$$

Finally, we can find the time the object takes to reach the maximum height:

$$t=\dfrac{v_0}{g}=\dfrac{60\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=6.1\ s.$$

Answer:

$y_{max}=183.7\ m, t=6.1\ s.$