Question

Using function notation, how would we represent when an object hitsthe ground if time is measured in seconds and height is measured infeet?

Answer 1

The height h(t) of a projectile above the ground after t seconds is given by; (neglecting air resistance)

$h(t)\text{ = -(}\frac{1}{2})gt^2+v_{0_{}}t+h_0$

where;

$\begin{gathered} v_0\text{ = initial velocity} \\ h_0=\text{ initial height} \\ h\mleft(t\mright)=height\text{ at time t} \\ g\text{ = acceleration due to gravity} \\ t\text{ = time in seconds} \end{gathered}$

when the object hits the ground, the height h(t) will be equal to zero.

$h(t)\text{ = 0}$

so, from the height equation at time t when the height h(t) is zero, the equation becomes;

$\begin{gathered} h(t)\text{ = -(}\frac{1}{2})gt^2+v_{0_{}}t+h_0\text{ = }0 \\ \text{-(}\frac{1}{2})gt^2+v_{0_{}}t+h_0\text{ = }0\text{ }\ldots\ldots\ldots\ldots1 \end{gathered}$

The time when the object hits the ground can be represented by the equation 1 above.

By substituting the values of g,v and h respectively we can solve for t.

Assumming the object was dropped on a free fall from height h.

For free fall the initial velocity is zero.

$\begin{gathered} v_{0\text{ }}=0 \\ h_0\text{ = h} \end{gathered}$

Substituting this values into the equation 1 above, we have:

$\begin{gathered} \text{-(}\frac{1}{2})gt^2+(0)_{}t+\text{ h }=\text{ 0} \\ \text{-(}\frac{1}{2})gt^2+\text{ h }=\text{ 0} \end{gathered}$

Then solving for t, we have;

$\begin{gathered} \text{-(}\frac{1}{2})gt^2+\text{ h }=\text{ 0} \\ \text{-(}\frac{1}{2})gt^2=\text{ -h} \\ \text{divide through by }-(\frac{1}{2})g \\ t^2\text{ }=\text{ }\frac{-h}{-(\frac{1}{2})g} \\ t^2\text{ }=\frac{2h}{g} \\ \text{square root both sides;} \\ t\text{ =}\sqrt{\frac{2h}{g}} \end{gathered}$

Therefore, In function notation, the time t in seconds when an object dropped from height h(in feet) hits the ground is:

$t\text{ =}\sqrt{\frac{2h}{g}}$