Question

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?

Answer 1

Answer:

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:

        $y = v_{o}* t -\frac{1}{2} * g * t^{2} (1)$

• 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:

        $y = y_{o} - \frac{1}2}*g*t^{2} (2)$

• 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:

        $y -y_{o} = -\frac{1}{2}*g* t^{2} (3)$

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

       $y = v_{o}*t + (y- y_{o})$

       ⇒ $t =\frac{y_{o} }{v_{o} } =\frac{14 m}{20.6 m/s} = 0.68 s$

       ⇒ t = 0.68s