Question

a criket ball is dropped from the height of 80 m /sec . calculate the speed of ball when it hitts the ground ?​

Answer 1

Answer:

$\boxed{\mathfrak{Speed \ of \ ball \ when \ it \ hitts \ the \ ground \ (v) = 40 \ m/s}}$

Explanation:

From equation of motion we have:

$\boxed{ \bold{ {v}^{2} = {u}^{2} + 2gh}}$

g → Acceleration due to gravity (10 m/s²)

v → Final speed

According to the question,

Height (h) = 80 m

Initial speed (u) = 0 m/s (Dropped)

By substituting values in the equation we get:

$\rm \implies {v}^{2} = {0}^{2} + 2 \times 10 \times 80 \\ \\ \rm \implies {v}^{2} = 1600 \\ \\ \rm \implies v = \sqrt{ {40}^{2} } \\ \\ \rm \implies v = 40 \: m {s}^{ - 1}$

Answer 2

Answer :

➥ The speed of ball when it hits the ground = 39.59 m/s

Given :

➤ A criket ball is dropped from the height (h) = 80 m/s

To Find :

➤ Speed of the ball when it hits the ground = ?

Solution :

⚡ In the above question given that the height is 80 m/s, Intial velocity is 0 m/a because the cricket ball is dropped, and it's acceleration due to gravity is 9.8 m/s.

Now, we have mass of the Height, Intial velocity, acceleration due to gravity,

• Height (h) = 80 m/s
• Intial velocity (u) = 0 m/s
• Acceleration due to gravity (g) = 9.8 m/s²
• Final velocity (v) = ?

We can find the final velocity by using the third equation of motion which says v² = u² + 2gh.

☃️ So let's find final velocity (v) !

→ v² = u² + 2gh

→ v² = 0² + 2 × 9.8 × 80

→ v² = 0 + 19.6 × 80

→ v² = 0 + 1568

→ v² = 1568

→ v = $\sf{\sqrt{1568}}$

→ v = 39.59 m/s

║Hence, the speed of ball when it hits the ground is 39.59 m/s.║

$\:$

Some related equations :

⪼ v = u + gt

⪼ h = ut + ½ gt²

⪼ v² = u² + 2gh

Where,

• v is the final velocity.
• u is the intial velocity.
• g is the acceleration due to gravity.
• t is the time taken.
• h is the height.

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