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# A ball is dropped from the top of a tower of height 100m. Simultaneously, another ball was thrown upward from the bottom of the tower with a speed of 50\$\dfrac{m}{s}\$ (g=10m\${s^2}\$ ). These two balls would cross each other after a time:A) 1 secondB) 2 secondC) 3 secondD) 4 second

Last updated date: 11th Jun 2024
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Hint: In order to find the solution of the question we will use the distance formula of Newton which says that s=ut+\$\dfrac{1}{2}\$ \$a{t^2}\$, where s is the distance, u is the initial velocity of object, a is the acceleration, and t is the time. We will use acceleration due to gravity as ball is falling under gravity. They both will reach at the same time to X position and through this we will calculate time.

Step 1:
To understand the question we have a diagram. Have a close look.

Here in the image we can see a tower is there of 100m and a ball from 100m height is falling under gravity which is 10\$\dfrac{m}{{{s^2}}}\$ and another ball is there at the bottom of tower and thrown upward with a velocity of 50\$\dfrac{m}{s}\$. We need to find the time when they will meet each other.
Let say, they will meet at position having dotted lines in the diagram.
The initial speed of upper ball will be zero before falling which means u=0
The initial speed of the ball on the foot of the tower will be u=50\$\dfrac{m}{s}\$
Acceleration due to gravity will be 10\$\dfrac{m}{{{s^2}}}\$
Step 2:
Now, from the equation of law of motion we have s=ut+\$\dfrac{1}{2}\$ \$a{t^2}\$, where s is the distance, u is the initial velocity of object, a is the acceleration, and t is the time.
For ball 1, s=h1(say)
Then h1=0(t) + \$\dfrac{1}{2}\$ (10)\${t^2}\$ \$ \Rightarrow \$ h1=\$5{t^2}\$ ……. (1)
For ball 2 s=h2(say)
Then h2=50t + \$\dfrac{1}{2}\$ (10)\${t^2}\$ \$ \Rightarrow \$ h2=50t−\$5{t^2}\$ ……. (2)
In above equation we have used the values which is stated above in step 2
The total height of the tower is 100m then we can say that h1+h2 equal to height of the tower.
Then adding equation (1) and (2), h1+h2= \$5{t^2}\$+50t−\$5{t^2}\$
This will give us height of tower H=50t
Evaluating the t with the total height, 100m=50t
This implies t =2 sec
These two balls would cross each other after a time: 2sec
Hence option B is correct.

Note: This problem can be solved using relative velocities or basically fixing a frame of reference to one of the balls and basically then one will have to travel the entire 100m distance wrt the other.