Question

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

Answer 1

Explanation:

use the formula

$\sqrt{2 \times 9.8 \times 19.6}$

Answer 2

Answer

Given :-

• Initial veloicty, u = 0 m/s
• Final velocity, 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
• g = 9.8 m/s²

Substituting the value in 3rd equation of motion :-

$\boxed{\sf v^2 = u^2 + 2as}$

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

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

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

$\sf v = 19.6 m/s$

$\boxed{\underline{\mathrm\red{Final \:velocity = 19.6 m/s}}}$