Question

Please Help

The expression represents the distance in feet an object falls after seconds. The object is dropped from a height of 906 feet.


What is the height in feet of the object 2 seconds after it is dropped?


Write an expression representing the height of the object in feet seconds after it is dropped.

Answer 1

Answer:

$842\ \text{ft}$

$y(t)=906-16t^2$

Step-by-step explanation:

Let $y$ be the height the object from the ground.

$s$ be the initial height of the object = 906 ft

$u$ = Initial velocity = 0

$g$ = Acceleration due to gravity = $32\ \text{ft/s}^2$

$t$ = Time

The expression would be

$y(t)=s-ut-\dfrac{1}{2}gt^2\\\Rightarrow y(t)=906-16t^2$

The required expression is $y(t)=906-16t^2$

At $t=2\ \text{s}$

$y(2)=906-16\times 2^2\\\Rightarrow y(2)=842\ \text{ft}$

The height of the object after 2 seconds of falling is $842\ \text{ft}$.