Solving vertical motion problems

Vertical motion models the vertical flight of an object, either upward or downward

Vertical motion is any type of upwards or downwards motion that is constant. In a vertical motion problem, you may be asked about instantaneous velocity, and/or average velocity.

To solve for instantaneous velocity we will need to take the derivative of our position function.

The position function is $s(t)$ and the instantaneous velocity is $v(t)$ (which is $s'(t)$).

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To solve for average velocity we will need our general formula

$\Delta{v(t_1t_2)}=\frac{s(t_2)-s(t_1)}{t_2-t_1}$

Remember, for instantaneous velocity we only require one time point but for average velocity we require two time points.

Vertical motion problems where an object is thrown up from the ground

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Dropping an object from the top of a building

Example

A watermelon is dropped from the top of a building that’s $28$ meters high. The position function of the watermelon is given by

$s(t)=-4t^2+28$

where $s$ is measured in meters, and $t$ is measured in seconds.

1. What is the instantaneous velocity of the watermelon when $t=4$?

2. When will the watermelon hit the ground?

3. What is the average velocity from $t=2$ until the watermelon hits the ground?

First we need to find the derivative of $s(t)$.

$s'(t)=v(t)=-8t$

To find velocity when $t=4$, we’ll plug $t=4$ into $v(t)$.

$v(4)=-8(4)$

$v(4)=-32$

Our velocity at $t=4$ is $-32$ m/s.

When the watermelon hits the ground, $s(t)=0$, because the position of the watermelon is at a height of $0$ meters off the ground. Therefore, we need to set $s(t)=-4t^2+28$ equal to $0$ and solve for $t$.

$-4t^2+28=0$

$-4t^2=-28$

$t^2=7$

$t=\sqrt{7}\approx 2.65$

The watermelon hits the ground at $t\approx 2.65$ s.

Vertical motion is any type of upwards or downwards motion that is constant.

To find average velocity, we’ll use the formula for average velocity,

$\Delta{v(t_1,t_2)}=\frac{s(t_2)-s(t_1)}{t_2-t_1}$

and remember that $t_1=2$ and $t_2=\sqrt7$.

$\Delta{v(t_1,t_2)}=\frac{-4\left(\sqrt7\right)^2+28-\left[-4(2)^2+28\right]}{\sqrt7-2}$

$\Delta{v(t_1,t_2)}=\frac{-4(7)+28-\left[-4(4)+28\right]}{\sqrt7-2}$

$\Delta{v(t_1,t_2)}=\frac{-28+28-\left[-16+28\right]}{\sqrt7-2}$

$\Delta{v(t_1,t_2)}=\frac{16-28}{\sqrt7-2}$

$\Delta{v(t_1,t_2)}=\frac{-12}{\sqrt7-2}\approx-18.58$

Average velocity from $t=2$ until the watermelon hits the ground is $-18.58$ m/s.

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