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

t= 0.68 s

Step-by-step 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 -(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) = -(1)/(2)*g* t^(2) (3)`

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

`y = v_(o)*t + (y- y_(o))`

⇒
`t =(y_(o) )/(v_(o) ) =(14 m)/(20.6 m/s) = 0.68 s`

⇒ t = 0.68s