Question

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?

Answer 1

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`.