A ball is thrown upward from the ground with an initial speed of 20.6 m/s; at the same instant, another ball is dropped from a building 14 m high. After how long will the balls be at the same height above the ground?

t= 0.68 s

Explanation:

Neglecting air resistance, both balls are only under the influence of gravity, so we can use the kinematic equation for vertical displacement for both balls.First of all, we define two perpendicular axes, coincidently with horizontal and vertical directions, that we denote as x-axis and y-axis respectively.Assuming that the upward direction is the positive one, g must be negative as it always points downward.Taking the ground as our zero reference for the vertical axis (y axis), the equation for the ball thrown upward can be written as follows:

       

As the second ball is dropped, its initial velocity is 0. Taking the height of the building as the initial vertical position (y₀), we can write the equation for the vertical displacement as follows:

       

As the left sides of  (1) and (2) are equal each other (the height  of both balls above the ground must be the same), the time must be the same also.We can rearrange (2) as follows:

       

Replacing the right side of (3) in (1), we get:

       

       ⇒

       ⇒ t = 0.68s

Multiply the 3 dimensions, assuming it is a rectangular cuboid, to get the volume in cm³ divide mass by volume to get density in g/cm³