A body is thrown vertically upward from a point 'A' 125 m above the ground. It goes up to a maximum height of 250 m above the ground and passes through 'A' on its downward journey. The velocity of the body when it is at a height of 70 m above the ground is (g = 10ms^–2)

Initial velocity(u) of the body is 60ms⁻¹

Distance(S) traversed by the body is (250−70)m

To find the velocity at the mentioned height, we use the formula of motion:
$$v^2=u^2+2gS$$…… (1)
Substituting the values in equation (1), we get:
$$\begin{gathered} v^2=u^2+2gS \\ \Rightarrow v^2=0^2+2\times 10\times \left(250-70\right) \\ \Rightarrow v^2=2\times 10\times 180 \end{gathered}$$

Simplifying the above equation, we get:

v2=3600

∴  v = 60ms⁻¹