Question

An object was thrown vertically upwards and it reached it peak height after 2.5 seconds before dropping back. What is the peak height measured from the throwing point? answer in meters

Answer 1

The maximum height

$$h_{max}=\frac{v_0^2}{2g}=\frac{gt^2}{2}$$ $$=\frac{9.8\:\rm m/s^2\times (2.5\: s)^2}{2}=30.6\:\rm m$$