Question

a boll falls of a table and rejest the ground in a 1 sec assuming 9= 10meter /s ​

Answer 1

Correct Question :-

• A ball falls off a table and reaches the ground in 1 s. Assuming g=10 m/s², calculate its speed on reaching the ground and the height of table.

SoluTion :-

Given :-

• Acceleration due to gravity (g) = 10 m/s
• Initial Velocity (u) = 0 m/s
• Time (t) = 11 s

To Find :-

• Final Velocity (v)
• Distance Covered (h)

Formula Used :-

• v = u + gt

→ v = 0 + 10 × 1

→ v = 0 + 10

→ v = 10

Thus, the final velocity is 10 m/s

So, the speed on reaching the ground and is 10 m/s.

Now, we will calculate Distance Covered

Formula Used :-

• v² - u² = 2gh

Subsituting Values

→ (10)² - (0)² = 2 × 10 × h

→ 100 - 0 = 20 × h

→ 100 = 20 h

→ 20 h = 100

→ h = 100/20

→ h = 5

So, the distance covered is 5 m

So, the height is 5 m

Answer 2

Correct Question :-

A ball falls off a table and reaches the ground in 1 s . Assuming g = 10 m/s^2 , calculate its speed on reaching the ground and the height of table.

Answer :-

Given :-

• Initial velocity, u = 0 m/s
• Time taken, t = 1 sec
• Acceleration due to gravity, g = 10 m/s²

To find :-

• Final velocity - v
• Distance travelled - s

Solution :-

Calculating Final velocity :-

• u = 0 m/s
• t = 1 sec
• g = 10 m/s²

Substituting the value in 1st equation of motion :-

$\sf v = u + at$

$\sf v = 0 + 10 \times 1$

$\sf v = 10 m/s$

Final velocity = 10 m/s.

Calculating Distance travelled :-

• u = 0 m/s
• t = 1 sec
• g = 10 m/s²

Substituting the value in 2nd equation of motion :-

$\sf s = ut + 1/2 at^2$

$\sf s = 0 + 1/2 \times 10 \times 1^2$

$\sf s = 5 m$

Distance travelled = 5 m