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]
Answer: 4.52 seconds
Step-by-step explanation:
We know that acceleration is the rate of change of the velocity, and velocity is the rate of change of the position.
If we define h(t) as the position, then we have that
h'(t) = velocity
h''(t) = acceleration
And we know that the acceleration is equal to -9.8m/s^2
then we have:
h''(t) = -9.8m/s^2 and the initial condition h(0) = 100m
For solving this, we integrate two times with respect to the time:
h'(t) = (-9.8m/s^2)*t + h'(0)
where h'(0) is the initial velocity, that in this case is 0m/s because the ball is droped.
We integrate again:
h(t) = ((-9.8m/s^2)/2)*t^2 + h(0) = -4.9m/s^2*t^2 + 100m
we need to find the time t in which h(t) = 0
0 = -4.9m/s^2*t^2 + 100m
4.9m/s^2*t^2 = 100m
t = (√100/4.9)s = 4.52 seconds
