A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50 meter/sec. from the surface of the tower, then they will meet at which height from the surface of the tower

The correct answer is 80 meters.

Let us solve the problem step-by-step.

Step 1: Understand the motion of both balls
Let the tower have a height of $H=400\text{meters}$.
Let the ground (surface of the tower) be the reference level where height $y=0$.
The first ball (Ball 1) is dropped from the roof of the tower ($y=400\text{m}$) with an initial velocity $u_1=0\text{m/s}$.
The second ball (Ball 2) is thrown upwards from the ground ($y=0\text{m}$) with an initial velocity $u_2=50\text{m/s}$.
Let $g$ be the acceleration due to gravity, directed downwards. We will use $g=10{\text{m/s}}^2$ for calculations.

Step 2: Write equations of motion for both balls
Let $t$ be the time (in seconds) after they start when they meet.
At time $t$, the height of Ball 1 from the ground is given by:

$y_1=H-\frac{1}{2}g{t}^2$

$y_1=400-5t^2$

At time $t$, the height of Ball 2 from the ground is given by:

$y_2=u_{2}t-\frac{1}{2}g{t}^2$

$y_2=50t-5t^2$

Step 3: Find the time when they meet
When the two balls meet, their heights are equal: $y_1=y_2$.

$400-5t^2=50t-5t^2$

By cancelling $-5t^2$ from both sides, we get:

$400=50t$

$t=\frac{400}{50}=8\text{seconds}$

Step 4: Find the height from the surface of the tower at which they meet
To find the meeting height, substitute $t=8$ into the equation for $y_2$:

$y_2=50\left(8\right)-5\left(8^2\right)$

$y_2=400-5\left(64\right)$

$y_2=400-320=80\text{meters}$

Thus, the two balls will meet at a height of 80 meters from the ground surface of the tower.