An object is said to be a freely falling body if it is thrown or dropped vertically at an initial velocity and then only gravity affects its motion.

The position of any freely falling body is determined by the initial velocity and the initial height.

If *h* is the height measured in feet, *t* is the number of seconds the object has fallen from an initial height
${h}_{0}$
with an initial velocity or speed
${v}_{0}$
(inft/sec), then the model for height of a falling object is:
$h\left(t\right)=-16{t}^{2}+{v}_{0}t+{h}_{0}$

The “–16*t*^{2}” term comes from the acceleration due to gravity pulling the object towards the ground.
The velocity of the object at a particular time t is given by:
$v\left(t\right)=-32t+{v}_{0}$

When an object is thrown upwards from ground with a particular initial velocity, the initial height is zero and when an object is dropped from an initial height the initial velocity is zero. If the value of h is in meters and s in meters/sec, the acceleration due to gravity in meters/sec is 4.9, the equation becomes: $h\left(t\right)=-4.9{t}^{2}+{v}_{0}t+{h}_{0}$ and then the velocity of the object at a particular time t is given by: $v\left(t\right)=-9.8t+{v}_{0}$

**Example 1:**

A ball is thrown vertically upward with an initial speed of 80 ft/sec. How high will the ball be after 3 seconds?

*t* = 3 and
${v}_{0}$
= 80 ft/sec

So, *h* = –16(3)^{2} + 80(3)
= –144 + 240
= 96 feet

**Example 2:**

An object is dropped from a height of 120 feet. Assuming that there is no air resistance, how long does it takes to reach the ground?

If *h* is measured in feet, *t* is the number of seconds the object had fallen, and
${h}_{0}$
is the initial height from which the object was dropped, then the model for the height of falling object is:

$h=-16{t}^{2}+{h}_{0}$

Note that the initial velocity is zero here.

Substitute 0 for *h* and 120 for
${h}_{0}$
in the model.
$0=-16{t}^{2}+120$

Solve the equation for *t*.
$\begin{array}{l}0-120=-16{t}^{2}+120-120\\ -120=-16{t}^{2}\\ \frac{-120}{-16}=\frac{-16{t}^{2}}{-16}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7.5={t}^{2}\end{array}$

Taking the square root: $\begin{array}{l}t=\sqrt{7.5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx \pm 2.74\end{array}$

Since *t* represents time, it cannot be negative.

Therefore, the object will reach the ground in about 2.74 seconds.

These equations are simplified. They ignore air resistance and the gravitational constant is approximate. Also, this model only works for the surface of the Earth (at sea level). The model on other planets will be different because their gravity is different. For example, on the surface of the moon, with h in meters and ${v}_{0}$ in m/sec, the falling object model is $h=-0.8{t}^{2}+{v}_{0}t$ + ${h}_{0}$ .