Question

A ball is gently dropped from a height of 20m . If it's velocity increases uniformly

Answer 1

$\huge\star\sf\underline\pink{Solution :-}$

We have,

$\sf{ Initial\: velocity (u) = 0} \\ \\ \sf{ Acceleration (a) = 10\:ms^{-1}} \\ \\ \sf{ Distance (s) = 20\:m}$

$\sf\underline\red{To\:Find:-} \\ \\ \: \: \: \: \: \bullet\sf{Final\: velocity (v) = ?}$

$\sf\underline\red{Formula\:used:} \\ \\ \pink{\boxed{\sf{ v^2 = u^2 + 2as}}}$

Now,

We can put values in the Formulas,

$\implies\sf{ v^2 = (0)^2 + 2 \times 10 \times 20} \\ \\ \\ \implies\sf{ v^2 = 0 + 400} \\ \\ \\ \implies\sf{ v = \sqrt{400} } \\ \\ \\ \implies\sf{ v = 20\:ms^{-1} }$

Thus,

The ball is strike with velocity of 20 metres/second.

Let's calculate Time:

We know that,

$\huge\bullet\pink{\boxed{\sf{ v = u + at}}}$

So,

$\implies\sf{ 20 = 0 + 10 \times t} \\ \\ \\ \implies\sf{ 10t = 20} \\ \\ \\ \implies{\sf{ t = \dfrac{20}{10} }} \\ \\ \\ \implies\sf{ t = 2 \: s}$

Thus,

The ball will strike the ground after 2 seconds.