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

`842\ \text{ft}`

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

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-(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* 2^2\\\Rightarrow y(2)=842\ \text{ft}`

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