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?

1 Answer
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(t)# - Position of the object after time #t#,
#v_o# - Initial velocity of the object,
#v(t)# - 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(t)-X_o = v_ot+1/2 at^2# : displacement-time relation

Combining the above two relation one gets,

# v^2(X) = v_o^2+2a(X-X_o)# : velocity-displacement relation

This Problem : Given #v_o# and #a# and we are asked to determine the displacement #X(t)-X_o# at time #t=1.00s#.

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(t)-X_o=v_ot-1/2 g.t^2 = 19.6 m#

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