A ball is dropped from the top of a building of height $80m$. At the same instant another ball is thrown upwards with speed $50m/s$ from the bottom of the building. The time at which balls meet is
(A) $1.6s$
(B) $5s$
(C) $8s$
(D) $10s$

Hint The second equation of motion can be used to solve the given question. The height of the building is constant. Thus equating the distances covered by both balls can give the time at which they meet.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Where $s$ is the distance covered by the body.
$u$ is the initial velocity of the body.
$v$ is the final velocity of the body.
$a$ is the acceleration.
$t$ is the time taken.

Complete Step by step solution
It is given in the question that,
The height of the building, $h = 80m$
Ball A is dropped from the top of the building with initial velocity of, ${u_A} = 0m/s$
Ball B is thrown upwards from the bottom of the building with initial velocity of, ${u_B} = 50m/s$
We know that the only force acting here is gravity therefore the acceleration of both balls is $g = 9.8m/s$ in the downward direction.
When the balls meet, the sum of the distances traveled by them totals to the height of the building.
Let, distance traveled by ball A be, ${s_A}$.
And the distance traveled by ball B be, ${s_B}$.
Then, $h = {s_A} + {s_B}$
From the second equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
For ball A,
$u = 0$
$a = 9.8m{s^{ - 2}}$
Therefore,
${s_A} = 0 \times t + \dfrac{1}{2} \times 9.8 \times {t^2}$
${s_A} = \dfrac{1}{2} \times 9.8{t^2}$
For ball B,
$u = 50m$
$a = - 9.8m/{s^2}$ (it is negative because the gravity tries to retard the motion of the ball)
Therefore,
${s_B} = 50 \times t - \dfrac{1}{2} \times 9.8 \times {t^2}$
${s_B} = 50t - \dfrac{1}{2} \times 9.8 \times {t^2}$
Adding the distances covered by both balls,
$h = {s_A} + {s_B}$
$80 = \dfrac{1}{2} \times 9.8{t^2} + 50t - \dfrac{1}{2} \times 9.8{t^2}$
$80 = \dfrac{1}{2} \times 9.8{t^2} + 50t - \dfrac{1}{2} \times 9.8{t^2}$
$80 = 50t$
$t = \dfrac{{80}}{{50}} = 1.6$
The time when both balls meet is at$1.6\sec $.

Therefore option (1) is correct.

Note Velocity and acceleration are vector quantities. They can be positive or negative, depending on their directions. Velocity of an object may be negative when the object moves in the negative direction. But acceleration can be negative in two cases, first when the acceleration is in the opposite direction, or when the object is retarding.