Question

a stone is dropped from a top of a building which is 19.6 m high. find the velocity with which it hits the ground.​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Final\:velocity=14\sqrt{2}\:m/s}}$

${\bold{\underline{\underline{Step-by-step\:explanatiom:}}}}$

• In the given question information given about a stone is dropped from a top of a building which is 19.6 m high.

• We have to find the velocity with which it hits the ground.

$\underline \bold{given : } \\ \implies Initial \: velocity(u) = 0 \: m/s \\ \\ \implies Distance(s) = 19.6 \: m \\ \\ \implies Acceleration(a) = 10 \: m/ {s}^{2} \\ \\ \underline \bold{To \: Find : } \\ \implies Final \: velocity(v) = ?$

• According to given question :

$\bold{Using \: second \:equation \: of \: motion : } \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {v}^{2} = {0}^{2} + 2 \times 10 \times 19.6 \\ \\ \implies {v}^{2} = 392 \\ \\ \implies v = \sqrt{392} \\ \\ \bold{\implies v = 14 \sqrt{2} \: m/s}$

Answer 2

$\huge\sf{Answer:-}$

• 0 m/s is the initial velocity
• 19.6m is the distance
• 10 m/s is the acceleration.

Given question

A stone is dropped from a top of a building which is 19.6 m high.We have to find the final velocity.

So,

According to the given question

= v² = u² + 2as

= v² = 0² + 2 × 10 × 19.6

= v² = 392

$= \sqrt{392} \\ = v = 14 \sqrt{2} m/s$