Question

A body is released from rhe top of height 19.6m. Find the velocity of the body

Answer 1

Answer: 19.6 m/s

Explanation:

we know,

v^2-u^2 = 2as

2as= 2×9.8×19.6 (since a= 9.8, s=19.6)

=19.6×19.6= 19.6^2

Therefore

v^2-u^2=19.6^2

Since it's free fall, u=0 therefore u^2=0

Therefore, v^2 =19.6^2

V=19.6 m/s

Answer 2

we know that gravitational acceleration = 9.8m/s²

given;

• h = 19.6m
• u = 0m/s (as its dropped)
• v = ?

Thus by 3rd equation of motion;

• v² = u² + 2gh
• v² = 0² + 2 × 196/10 × 98/10
• v² = 2×98×196/100
• v² = 196²/10²
• v = √{196²/10²)
• v = 196/10
• v = 19.6 m/s

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