A body is thrown vertically upward with initial velocity

A body is thrown vertically upward with initial velocity $v_0$, and (neglecting air resistance) its velocity at time $t$ is $v=v_0-gt$. Find its height above the launch point $t$ seconds after launch.

Step 1: Integrate velocity to get height

\[h(t)=\int(v_0-gt)\,dt=v_0t-\frac{gt^2}{2}+C.\]

Step 2: Apply the initial condition

At $t=0$ the body is at the launch point, $h(0)=0$, which forces $C=0$.

\[\boxed{h(t)=v_0t-\frac{1}{2}gt^2}\]

Step 3: Maximum height

The maximum occurs when $v=0$, i.e. $t_{\max}=v_0/g$, giving

\[h_{\max}=v_0\cdot\frac{v_0}{g}-\frac{1}{2}g\left(\frac{v_0}{g}\right)^2=\frac{v_0^2}{2g},\]

which also follows directly from energy conservation ($\tfrac12mv_0^2=mgh_{\max}$).

Step 4: Total flight time

Setting $h=0$ gives $t\left(v_0-\tfrac12gt\right)=0$, so besides $t=0$ the body lands again at

\[t_{\text{total}}=\frac{2v_0}{g},\]

exactly twice the time to reach the top, reflecting the symmetry of motion under constant acceleration.