Question

Student drops a boll from the top of tower of height 19.6m . Find the velocity with which the ball hits the ground ?

Answer 1

Given :-

• Height (h) = 19.6 m
• Acceleration due to gravity (g) = 10 N (approx 9.8)
• Initial Velocity (u) = 0 m/s

To Find :-

• Final Velocity (v)

SoluTion :-

Formula Used :-

• v² - u² = 2gh (2nd equation of Motion)

Subsituting Values

→ v² - 0² = 2 × 10 × 19.6

→ v² = 19.6 × 19.6

→ v = √19.6 × 19.6

→ v = 19.6

Thus, the velocity at which the ball reaches the ground is 19.6 m/s

3 Equations of Motion :-

• v = u + at
• v² - u² = 2 as
• s = ut + ½ at²

Answer 2

Answer :-

Given :-

• Initial velocity, u = 0 m/s
• Height, h = 19.6 m
• Acceleration due to gravity, g = 9.8 m/s²

To find :-

• Final velocity - v

Solution

• u = 0 m/s
• h = 19.6 m
• a = g = 9.8 m/s²

Substituting the value in 3rd equation of motion :-

$\sf v^2 = u^2 + 2as$

$\sf v^2 = 0 + 2 \times 9.8 \times 19.6$

$\sf v^2 = 19.6 \times 19.6$

$\sf v = { (\sqrt{19.6} )}^{2}$

$\sf v = 19.6 m/s$