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

Answer:

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 = $V_{i}$t + $\frac{1}{2}$ $a_{y}$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 + $\frac{1}{2}$ $a_{y}$t²

-( 18 m - h )  = 0 - $\frac{1}{2}$gt²

-18 m + h =  - $\frac{1}{2}$gt²

h = 18 m - $\frac{1}{2}$gt² --------- let this be equ1

Now lets apply kinematic relation to the raising ball

h = (21.4 m/s)t - $\frac{1}{2}$gt²   ----------- let this be equ 2

so equ1 = equ2

18 m - $\frac{1}{2}$gt² = (21.4 m/s)t - $\frac{1}{2}$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