Question

A ball is thrown upward from the ground with an initial speed of 21.4 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground

Answer 1

after 0.8411 seconds, the balls are at the same height

Step-by-step explanation:

given the data in the question;

we take a look at the equation which relates the vertical displacement of the object, time elapsed, initial velocity and acceleration of the object;

Δy =
`V_(i)`t +
`(1)/(2)`
`a_(y)`t²

where Δy is the vertical displacement,
`V_(i)` is the initial velocity,
`a_(y)` is acceleration of the object and t is time elapsed.

Now let h represent the height above ground level where they meet.

so the distance moved by the falling ball will be;

18 m - h

The ball is moving freely under the influence of gravity.

so
`a_(y)` = -g

Using kinematic relation a freely falling ball

Δy =
`V_(i)`t +
`(1)/(2)`
`a_(y)`t²

-( 18 m - h ) = 0 -
`(1)/(2)`gt²

-18 m + h = -
`(1)/(2)`gt²

h = 18 m -
`(1)/(2)`gt² --------- let this be equ1

Now lets apply kinematic relation to the raising ball

h = (21.4 m/s)t -
`(1)/(2)`gt² ----------- let this be equ 2

so equ1 = equ2

18 m -
`(1)/(2)`gt² = (21.4 m/s)t -
`(1)/(2)`gt²

18 m = (21.4 m/s)t

t = 18 m / 21.4 m/s

t = 0.8411 s

Therefore, after 0.8411 seconds, the balls are at the same height