Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at
the rate of 10 m s-2, with what velocity will it strike the ground ? After what time
will it strike the ground?​

Answer 1

Answer :-

Given :-

• Height = 20 m
• Initial velocity = 0 ( because it is dropped )
• Acceleration = 10 m/s²

To Find :-

• Final velocity
• Time

Solution :-

Calculating final velocity :-

• h = 20 m
• u = 0
• a = 10 m/s²

Substituting the value in 3rd equation of motion :-

→ v² = u² + 2as

→ v² = 0 + 2 × 20 × 10

→ v² = 400

→ v = 20

Final velocity = 20 m/s

Calculating time :-

• v = 20 m/s
• u = 0
• a = 10 m/s²

Substituting the value in 1st equation of motion :-

→ v = u + at

→ 20 = 0 + 10t

→ 20 = 10t

→ t = 20 / 10

→ t = 2

Time = 2 seconds

Answer 2

Given :-

• Height = 20m
• Initial velocity = 0
• Acceleration = 10m/s²

To find :-

• Final velocity
• Find Time

Solution :-

Calculation :

• H = 20m
• U = 0
• A = 10m/s²

• Substituting the value,

Third equation of motion :

:$\implies$v² = u² + 2as

$~~~~~$:$\implies$v² = 0 + 2 × 20 × 10

$~~~~~~~~~~$:$\implies$v² = 400

$~~~~~~~~~~~~~~~$:$\implies$${\underline{\boxed{\purple{\frak{v~=~20}}}}}$★

$\therefore$ Hence,

• The Final velocity is = $\large\underline{\rm{20~m/s}}$

V E R I F I C A T I O N :

• V = 20m/s
• U = 0
• A = 10m/s

• Substituting the value,

First equation of motion :

:$\implies$v = u + at

$~~~~~$:$\implies$20 = 0 + 10at

$~~~~~~~~~~$:$\implies$20 = 10at

$~~~~~~~~~~~~~~~$:$\implies$t = $\large{\sf{\frac{20}{10}}}$

$~~~~~~~~~~~~~~~~~~~~$:$\implies$t = ${\cancel{\dfrac{20}{10}}}$

$~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$${\underline{\boxed{\pink{\frak{t~=~2}}}}}$★

$\large\dag$ Hence Verified !!

• Time = $\large\underline{\rm{2~second}}$