Question

a stone dropped from a tree takes 2 s to reach the ground. Find its velocity on striking the ground​

Answer 1

Given :

Initial velocity of the stone is zero.

Time taken by the stone to reach the ground is 2s.

To find :

Velocity of stone when it reaches the ground.

Solution :

Velocity is the rate of displacement of a moving object over time.

• Velocity is vector quantity having both magnitude and direction
• SI unit of velocity is metre per second

When a body is projected vertically downwards with an initial velocity from a height h then a = g and s = h.

So, g = 10m/s²

As we are provided with initial velocity , gravity and time we can use first equation of motion.

Using first equation of motion .i.e.,

• v = u + gt

where,

v denotes final velocity

u denotes initial velocity

g denotes gravity

t denotes time

Substituting all the given values in the equation,

➠ v = u + gt

➠ v = (0) + (10)(2)

➠ v = 20m/s$.$

thus, its velocity on striking the ground is 20m/s.

Some related equation of motion :

1st equation of motion : v = u + at.

2nd equation of motion : s = ut + 1/2at².

3rd equation of motion : v² = u² - 2as.

Answer 2

$\huge{ \bf{AnSwEr :}}$

The time taken by the stone to reach the ground is 2s

To find the velocity of the stone on reaching the ground

Here we can apply 1st equation of motion

➝ v = u + at

a = g (because it is projected vertically downwards)

so, g = 10m/s²

Initial velocity (u) = 0m/s

Substituting value in equation,

➝ v = 0 + (10)(2)

➝ v = 20m/s

Hence velocity of the stone on reaching the ground is 20m/s.