Question

A piece of stone is dropped from a tower 45m high while falling its uniform accelaration was 10m/s2.How long will it take to touch ground at the moment of touching the ground what would be its velocity?

Answer 1

hope u understand the calculations yourself

Answer 2

Given data :-

• A piece of stone is dropped from a tower 45 m high.

• Stone falling with uniform acceleration 10 m/s².

Here,

{ u = initial velocity, v = final velocity, a = acceleration of the particle, t = time taken by particle, s = displacement of the particle. }

→ Displacement of the stone ( s ) = 45 m

→ Acceleration of the stone ( a ) = 10 m/s²

Solution :-

Let, initial velocity of stone be zero.

Now {we use kinematical equation to find time taken by stone to touch ground.}

→ s = ut + ½ at²

→ 45 = 0 × t + ½ × 10 × t²

→ 45 = 5 × t²

→ 45 = 5 × t² i.e.

→ t² = 45/5

→ t² = 9

→ t = √9

→ t = 3 sec

Now, {we use kinematical equation to find velocity of stone at the moment of touching the ground.}

→ v = u + at

→ v = 0 + 10 × 3

→ v = 30 m/s

Hence, stone take 3 sec to touch ground at the moment of touching the ground, velocity of the stone is 30 m/s .

More info :

Kinematical equation for uniformly accelerated motion of particle.

→ v = u + at

→ s = ut + ½ at²

→ v² = u² + 2as

{ Generally we take g = 9.8 m/s² becasue falling body accelerated due to gravity. }