# If a rock is thrown upward with an initial velocity of 24.5 m/s where the downward acceleration due to gravity is 9.81 m/s2. what is the rock's displacement after 1.00 s?

Dec 18, 2014

Solving this problem requires knowing the Galilean Equations of Motion for the case of constant acceleration.

${X}_{o}$ - Initial position of the moving object,
$X \left(t\right)$ - Position of the object after time $t$,
${v}_{o}$ - Initial velocity of the object,
$v \left(t\right)$ - Velocity of the object after time $t$,
$a$ - Acceleration (a constant)

The Galilean Equations of Motion :

v(t) = v_o + at; : velocity-time relation,
$X \left(t\right) - {X}_{o} = {v}_{o} t + \frac{1}{2} a {t}^{2}$ : displacement-time relation

Combining the above two relation one gets,

${v}^{2} \left(X\right) = {v}_{o}^{2} + 2 a \left(X - {X}_{o}\right)$ : velocity-displacement relation

This Problem : Given ${v}_{o}$ and $a$ and we are asked to determine the displacement $X \left(t\right) - {X}_{o}$ at time $t = 1.00 s$.

Given v_o=24.5ms^{-1}, \qquad a=-g=-9.8ms^{-2}; \qquad t=1.00s

For this purpose the displacement-time relation is useful.

$X \left(t\right) - {X}_{o} = {v}_{o} t - \frac{1}{2} g . {t}^{2} = 19.6 m$

The displacement after $1.00 s$ is $19.6 m$.