Question

stone dropped from the top of a tower reaches the ground in 6 s. The height of the tower is (take g=10m/s²) *​

Answer 1

Given :-

In this question we are given that a stone is dropped from some height and reaches the ground in 6s and we know that there is an gravitational force acting in the downward direction and an air resistive force is acting on the particle in upward direction.

Time = t = 6 s

Acceleration due to gravity = g = 10 ms-²

To find :-

Height of the tower.

Let the height of tower = h, then second equation of Motion.

Sy = ut + 1/2 (gt²)

h = 0 + 1/2 × 10 × 6 × 6

h = 180 m

Using this value we can also find the velocity of particle.

v = √2gh

v = √2 × 10 × 180

v = √3600

v = 60 ms-¹

Hence,

The height of tower = h = 180 m

Velocity of particle while coming downward = v = 60 ms-¹.

Concept :-

Whenever a body is thrown from the top of a building or tower or whatever its initial velocity becomes zero and body tends to move with final velocity.

Answer 2

Given :

• Time taken by the stone to reach the ground = 6 s
• Initial velocity = 0 m/s
• Acceleration due to gravity = 10 m/s²

To find :

• Height of the tower = ?

Solution:

Let the height of the tower = s

Here, we will use the second equation of motion.

★ s = ut + 1/2 at²

Where,

• s = height of the tower
• u = Initial velocity
• t = time taken

Substituting the given values,

→ s = 0 × 6 + 1/2 × 10 × 6 × 6

→ s = 0 + 5 × 36

→ s = 5 × 36

→ s = 180

★ Height of the tower = 180 m

━━━━━━━━━━━━━━━━

Know More :

Equations of motion :

❖ First equation of motion - v = u + at

❖ Second equation of motion - s = ut + 1/2 at²

❖ Third equation of motion -v² - u² = 2as

Some definitions -

Displacement : It is the distance travelled by an object in a particular direction.

Acceleration : It is the rate of change of velocity of a body.

Velocity : It is distance travelled by a body in one second in a particular direction.