Question

An arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second. Let up be the positive direction. Because gravity is the force pulling the arrow down, the initial acceleration of the arrow is −9.8 meters per second squared.

Answer 1

Answer:

$s(t)=-9.8t^2+49t+58.8$

Step-by-step explanation:

We have been given that an arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second. Let up be the positive direction. Because gravity is the force pulling the arrow down, the initial acceleration of the arrow is −9.8 meters per second squared.

We know that equation of an object's height t seconds after the launch is in form $s(t)=-gt^2+v_0t+h_0$, where

g = Force of gravity,

$v_0$ = Initial velocity,

$h_0$ = Initial height.

For our given scenario $g=-9.8$, $v_0=49$ and $h_0=58.8$. Upon substituting these values in object's height function, we will get:

$s(t)=-9.8t^2+49t+58.8$

Therefore, the function for the height of the arrow would be $s(t)=-9.8t^2+49t+58.8$.