# If a ball is thrown vertically upward from the ground with a velocity of 144 ft/s, then its...

## Question:

If a ball is thrown vertically upward from the ground with a velocity of 144 ft/s, then its height after t seconds is {eq}s = 144t - 16t^2 {/eq}. What is the maximum height reached by the ball?

## Maxima and Minima

The principle of maxima and minima of any entity is determined by calculating the critical point for that quantity. The maximum value of the function is determined by substituting the critical value in the function.

Given Data

• The height of the ball is: {eq}s = 144t - 16{t^2} {/eq}

Differentiate the above equation with respect to {eq}t {/eq}.

{eq}\begin{align*} \dfrac{d}{{dt}}\left( s \right) &= \dfrac{d}{{dt}}\left( {144t - 16{t^2}} \right)\\ \dfrac{{ds}}{{dt}} &= 144 - 32t\cdots\cdots\rm{(I)} \end{align*} {/eq}

It is known that the first-order derivative of the height function with respect to the time is the velocity.

So, {eq}\dfrac{{ds}}{{dt}} = v {/eq}.

At the maximum height, the velocity is zero.

So, {eq}v = \dfrac{{ds}}{{dt}} = 0 {/eq}.

From equation (I),

{eq}\begin{align*} 0 &= 144 - 32t\\ t &= \dfrac{{144}}{{32}} \end{align*} {/eq}

Substitute the known values in the height function.

{eq}\begin{align*} s &= 144\left( {\dfrac{{144}}{{32}}} \right) - 16{\left( {\dfrac{{144}}{{32}}} \right)^2}\\ s &= {\left( {144} \right)^2}\left( {\dfrac{1}{{32}} - \dfrac{{16}}{{{{32}^2}}}} \right)\\ s &= 324\;{\rm{ft}} \end{align*} {/eq}

Thus, the maximum height reached by the ball is {eq}324\;{\rm{ft}} {/eq}.