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 25m/s (g=10ms2 ). These two balls would cross each other after a time: A) 1 second B) 2 second C) 3 second D) 4 second - Padhle.Online
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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 25m/s (g=10ms2 ). These two balls would cross each other after a time: A) 1 second B) 2 second C) 3 second D) 4 second

Here  we can see a tower is there of 100m and a ball from 100m height is falling under gravity which is 10m/s2 and another ball is there at the bottom of tower and thrown upward with a velocity of 25m/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 the 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 = 25m/s

Acceleration due to gravity will be 10ms2

Now, from the equation of law of motion we have s = ut+1/2 at2,

• where s is the distance
• u is the initial velocity of the object
• a is the acceleration
• t is the time.

For ball 1, s=h1

Then h1=0(t) + 12 (10)t2 ⇒  h1 = 5t2 ……. (1)

For ball 2 s=h2

Then h2=50t + 12 (10)t2 ⇒  h2 = 50t − 5t2 ……. (2)

In above equation we have used the values which is stated above in step

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= 5t2 + 25t−5t2

This will give us height of tower H= 25t

Evaluating the t with the total height, 100m=25t

This implies t =4 sec

These two balls would cross each other after a time: 4 sec

Principal, Babu Daudayal SVM, Mathura