Question

$\LARGE{\color{teal}{\textsf{\textbf{ Question }}}}$
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
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Answer 1

Answer:

u=0

u=0s=19.6m

u=0s=19.6ma=10m/s² ( acceleration due to gravity)

u=0s=19.6ma=10m/s² ( acceleration due to gravity)v=?

2×10×19.6=v²-0²

196×2=v²

√(196×2)=v

13√2=v

12×1.4{aprox}

16.8m/s= v

use 3d equation of motion

2as=v²-u²

Answer 2

Answer:

• Its final velocity before touching the ground is 19.8 m/s.

Explanation:

Given information,

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

• Initial velocity (u) = 0 m/s
• Acceleration = Acceleration due to gravity (g) = 10 m/s²
• Height of tower (h) = 19.6 m
• Final velocity (v) = ?

Using third eqⁿ of motion,

✪ v² = u² + 2as ✪

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance travelled

We have,

• u = 0 m/s
• a = g = 10 m/s²
• s = h = 19.6 m
• v = ?

Putting all values,

➻ v² = 0² + 2gh

➻ v² = (0 × 0) + 2(10)(19.6)

➻ v² = 0 + (2 × 10 × 19.6)

➻ v² = 2 × 10 × 19.6

➻ v² = 2 × 196

➻ v² = 392

➻ v = √(392)

➻ v = 19.79

➻ v ≈ 19.8

• Henceforth, its final velocity before touching the ground is 19.8 m/s.

Three equations of motion,

1. v = u + at
2. s = ut + ½ at²
3. v² = u² + 2as

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken
• s denotes distance travelled

Some important definitions,

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

• Distance travelled

The total path length covered by ans object is known as distance travelled.

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