Question

please answer the questions fast and correct​

Answer 1

Given

• A ball is dropped down
• Height = 20 m
• Acceleration = 10 m/s²

To Find

• Velocity of the body
• Time taken to strike the ground

Solution

● v²-u² = 2as [Third Equation of Motion]

● v = u + at [First Equation of Motion]

✭ Velocity of the body :

→ v²-u² = 2as

→ v²-0² = 2 × 10 × 20

→ v² = 400

→ v = √400

→ Velocity = 20 m/s

━━━━━━━━━━━━━━━━━━━━━

✭ Time taken :

→ v = u + at

→ 20 = 0 + 10 × t

→ 20 = 10t

→ 20/10 = t

→ Time = 2 sec

Answer 2

Answer:

Given :-

• Height when it dropped (H) = 20 m
• Acceleration = 10 m/s²

To Find :-

At what time it will hit ground

Solution :-

Firstly let's find its velocity by using newton third Equation

$\sf \: v {}^{2} = {u}^{2} + 2as$

$\sf \: v {}^{2} = 0 {}^{2} + 2(10)(20)$

$\sf \: v {}^{2} = 0 + 400$

$\sf \: v {}^{2} = 400$

$\sf \: v = \sqrt{400}$

$\sf \: v = 20$

Hence,

Final velocity of object is 20 m/s.

Let's find the time taken by using Newton first Equation

$\sf \: v = u + at$

$\sf 20 = 0 + 10 (t)$

$\sf \: 20 = 10t$

$\sf \: t = \dfrac{20}{10}$

$\sf \: t = 2 \: second$