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
Cosmic Defect
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.

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