Question

A body falls freely from rest, from a height of 25 m. Find

(a) The speed with which it strikes the ground.

(b) The time it takes to reach the ground.

Acceleration due to gravity at the surface of Earth is 10 m/s2

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Answer 1

Given:-

• Height ,h = 25m
• Initial velocity ,u = 0m/s
• Acceleration due to gravity ,g = 10m/s²

To Find :-

• Speed with which it strikes the ground
• Time taken to reach the ground

Solution :-

According to the Question

It is given that body falls from rest , from height of 25m

Firstly we will calculate the speed with which it strikes the ground .

For calculating the speed we will apply here kinematics equation

• v² = u² + 2gh

by putting the value we get

↠ v² = 0² + 2×(10)×25

↠ v² = 0 + 20×25

↠ v² = 500

↠ v = √500

↠ v = 22.36 m/s

• Hence, the speed with which it strikes the ground will be 22.36 m/s

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Now,

(b) The time it takes to reach the ground.

Again by using Kinematics Equation

• v = u + gt

by substituting the value we get

↠ 22.36 = 0 + 10 × t

↠ 22.36 = 10t

↠ t = 22.36/10

↠ t = 2.236 s

• Hence, the tíme taken to reach the ground will be 2.236 seconds .

Answer 2

Answer:

Given :-

• A body falls freely from rest, from a height of 25 m.
• Acceleration due to gravity at the surface of Earth is 10 m/s².

To Find :-

• (a) The speed with which it strikes the ground.
• (b) The time it takes to reach the ground.

Formula Used :-

$\clubsuit$ Third Equation Of Motion Formula :

$\longrightarrow \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2gh}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• h = Height

$\bigstar$ First Equation Of Motion Formula :

$\dashrightarrow \sf\boxed{\bold{\pink{v =\: u + gt}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• t = Time Taken

Solution :-

First, we have to find the speed with which it strikes the ground :

Given :

• Initial Velocity = 0 m/s²
• Acceleration due to gravity = 10 m/s²
• Height = 25 m

According to the question by using the formula we get,

$\mapsto \bf v^2 - u^2 =\: 2gh$

$\mapsto \sf v^2 - (0)^2 =\: 2 \times 10 \times 25$

$\mapsto \sf v^2 - (0 \times 0) =\: 20 \times 25$

$\mapsto \sf v^2 - 0 =\: 500$

$\mapsto \sf v^2 =\: 500 + 0$

$\mapsto \sf v^2 =\: 500$

$\mapsto \sf v =\: \sqrt{500}$

$\mapsto \sf\bold{\purple{v =\: 22.36\: m/s}}$

$\therefore$ The speed with which it strikes the ground is 22.36 m/s .

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Now, we have to find the time it takes to reach the ground :

Given :

• Final Velocity = 22.36 m/s
• Initial Velocity = 0 m/s
• Acceleration due to gravity = 10 m/s

According to the question by using the formula we get,

$\implies \bf v =\: u + gt$

$\implies \sf 22.36 =\: 0 + 10 \times t$

$\implies \sf 22.36 - 0 =\: 10t$

$\implies \sf 22.36 =\: 10t$

$\implies \sf \dfrac{22.36}{10} =\: t$

$\implies \sf 2.236 =\: t$

$\implies \sf\bold{\red{t =\: 2.236\: seconds}}$

$\therefore$ The time it takes to reach the ground is 2.236 seconds .