1 Answer

Anurag Prashant · Answer 1 · 2022-10-05T01:53:07+0000

Answer:

$842\ \text{ft}$

y(t)=906-16t^2

Explanation:

Let
be the height the object from the ground.

be the initial height of the object = 906 ft

= Initial velocity = 0

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

= Time

The expression would be

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

The required expression is

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
$842\ \text{ft}$ .

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