physics question

Question

Bill kicks a rock off the top of his apartment building. It strikes a window across the street 14 m away. The acceleration of gravity is 9.8 m/s 2 . If the window is 18 m below the position where Bill contacted the rock, how fast was it moving when it left Bill’s foot? Answer in units of m/s.

Answer 1

Start with your general motion equation - you will solve it in both x and y:

d = d0 + v0 t + (1/2) a t^2; d is the displacement, v is the velocity, a is the acceleration, t is the time, and subscript 0 indicates at time t = 0.

Because he kicked the rock off the top of the building, assume that it started out traveling horizontally only. So v0y = 0 and v0x = v0

Take the origin to be at the point the rock left the building; x0 = y0 = 0

This is a free-fall problem, so ax = 0 and ay = a = g = 9.8 m/s downward

y = 18 m (given) and x = 14 m (given)

I am defining positive as the toward the other building (direction of initial motion) and down.

Motion problems (any vector problems) are always solved separately in x and y. The reduced equations are: x = v0 t and y = (1/2) g t^2 The x equation has the velocity needed, but depends on time; so the y equation must be solved to get time.

==> t = sqrt[ (2 y) / g] = sqrt[ 2(18m) / (9.8 m/s^2) ] = 3.67 s (3.7 s)

Solve x equation for v0: ==> v0 = x/t = (14m) / (3.67 s) = 3.8 m/s