An object is thrown vertically upwards with an initial velocity, $u$ of $20m/s$. The motion of the object follows the differential equation $\frac{ds}{dt}=u-gt$, where $s$ is the height of the object in meters at time $t$ seconds and $g=9.8m/s^2$. Determine the height of the object after $3$ seconds if $s=0$ when $t=0$.



This is a first order differentiatial equation, and to determine the height of the object at the given conditions, then when have to obtain the general solution and particular solution then apply the boundary conditions.
$\frac{ds}{dt}=u-gt\Rightarrow\int{ds}=\int(u-gt)dt$
$s=ut-\frac{gt^2}{2}+c$ this is the general solution. Substitute the conditions at $s=0,t=0$
$0=u(0)-\frac{0^2}{2}+c$
$c=0$
The particular solution now is:
$s=ut-\frac{gt^2}{2}+0$
Substitute $u=20m/s$ and $g=9.8m/s^2$ and $t=3sec$
$s=20(3)-\frac{9.8(3)^2}{2}=60-44.1=15.9$