Problem 4 on Motion in a Straight Line

Description:

Problem

A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the position of 3^rd, 4^th and 5^th ball when the 6^th ball is being dropped.

Solution

Let’s discuss the motion path of the balls first.

Motion path of the balls

The sign convention chosen for this problem is −

• Down is positive
• Up is negative

All balls start their motion from rest. Hence, initial velocity is zero for all of them. (u = 0)

Because the person drops a ball every second, it is clear that when the 6th ball is dropped,

• 3^rd ball has already travelled for 3 seconds.
• 4^th has already travelled for 2 seconds.
• 5^th ball has already travelled for 1 second.

Let the displacement of n^th ball be x_n.

Calculating position of 3^rd ball

Initial Velocity of each ball, u = 0 m/s

Acceleration, g = +10 m/s²

Time for which 3^rd ball has travelled = 3 s

Taking roof-top point as the origin, displacement of 3^rd ball, S₃ is,

S = ut + 1/2 at²

S₃ = 0 × (3) + 1/2(9.8) × (3)²

S₃ = 44.1 m

Calculating position of 4^th ball

Initial Velocity of each ball, u = 0 m/s

Acceleration, g = +10 m/s²

Time for which 3^rd ball has travelled = 2 s

Taking roof-top point as the origin, displacement of 4^th ball, S₄ is,

S = ut + 1/2 at²

S₄ = 0 × (2) + 1/2(9.8) × (2)²

S₄ = 19.6 m

Calculating position of 5^th ball

Initial Velocity of each ball, u = 0 m/s

Acceleration, g = +10 m/s²

Time for which 3^rd ball has travelled = 1 s

Taking roof-top point as the origin, displacement of 5^th ball, S₅ is,

S = ut + 1/2 at²

S₅ = 0 × (1) + 1/2(9.8) × (1)²

S₅ = 4.9 m