Question

how long will a stone take to fall from the top of a building 80 metre high and what will be the velocity of the stone or reaching on the ground (
$<marquee \: h$
$g \: = 10 \: m \: per \: {s}^{2}$
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Answer 1

Answer:

Time = 4 seconds

Final velocity = 40 m/s

Explanation:

Given :

Height of the building = s = 80 metres

Gravitational force = g = 10 m/s²

Initial velocity = u = 0 m/s (The stone starts from rest)

To find :

• Time taken
• Speed of the stone when it reaches down

Using second equation of motion :

$s=ut+\frac{1}{2}at^{2}$

Here a = g,

$80 = 0 \times t + \frac{1}{2} \times 10t^{2}$

$80=5t^{2}$

t²=16

t = 4 seconds

The time taken for the stone to reach the ground is 4 seconds

Using first equation of motion :

V=u+at

V=0+10×4

V=0+40

V=40 m/s

The velocity of the stone on reaching the ground is equal to 40 m/s

Answer 2

GIVEN:

• Height of building = 80 m
• Acceleration = 10 m/s²
• Initial velocity = 0 m/s

TO FIND:

• Time taken
• Speed of stone

SOLUTION:

Finding time using second equation of motion

s = ut + ½ at²

Where

• s : distance = 80m
• u : initial velocity = 0 m/s
• a : acceleration = 10 m/s²
• t : time = ?

Substituting the values we have

→ 80 = 0(t) + ½ × ( 10 )( t² )

→ 80 × 2 = 10 t²

→ 160/10 = t²

→ 16 = t²

→ t = √16 = 4 s

Using , the first equation of motion

v = u + at

Where

• v : final velocity = ?
• u : initial velocity = 0 m/s
• a : acceleration = 10 m/s²
• t : time = 4 s

Substituting the values we have

→ v = 0 + 10(4)

→ v = 40 m/s

Hence, the velocity when the stone will reach the ground = 40 m/s