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

after 0.8411 seconds, the balls are at the same height

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 = t + t²

where Δy is the vertical displacement, is the initial velocity, 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 = -g

Using kinematic relation a freely falling ball

Δy = t + t²

-( 18 m - h )  = 0 - gt²

-18 m + h =  - gt²

h = 18 m - gt² --------- let this be equ1

Now lets apply kinematic relation to the raising ball

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

so equ1 = equ2

18 m - gt² = (21.4 m/s)t - 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