Exercise 1.1.11: A dropped ball accelerates downwards at a constant rate 9.8 meters per second squared. Set up the differential equation for the height above ground h in meters. Then supposing h(0) = 100 meters, how long does it take for the ball to hit the ground.

The ball's acceleration is the rate of change of its velocity, and its velocity is the rate of change of its height. So if [tex]h(t)[/tex] is the ball's height at time [tex]t[/tex], then[tex]h''(t)=-9.8\dfrac{\rm m}{\mathrm s^2}[/tex]The ball is dropped from [tex]h(0)=100\,\mathrm m[/tex], and since it's being dropped there is no additional velocity added to it, so [tex]h'(0)=0\dfrac{\rm m}{\rm s}[/tex].Integrating both sides of the ODE gives[tex]\displaystyle\int h''(t)\,\mathrm dt=h'(t)=\left(-9.8\dfrac{\rm m}{\mathrm s^2}\right)t+C_1[/tex]and since [tex]h'(0)=0[/tex], we have [tex]C_1=0[/tex]. Integrating again gives[tex]\displaystyle\int h'(t)\,\mathrm dt=h(t)=\dfrac12\left(-9.8\dfrac{\rm m}{\mathrm s^2}\right)t^2+C_2[/tex]and since [tex]h(0)=100\,\mathrm m[/tex], we get [tex]100\,\mathrm m=C_2[/tex]So the ball's height at any time is governed by[tex]h(t)=100\,\mathrm m+\left(-4.9\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]The ball hits the ground when [tex]h(t)=0[/tex]:[tex]100\,\mathrm m+\left(-4.9\dfrac{\rm m}{\mathrm s^2}\right)t^2=100\implies t=\approx4.5\,\mathrm s[/tex]