Physics, asked by s13968, 5 months ago

A marble is dropped from the top of a tower. What distance does it fall after 3s of
dropping? (g=9.8m/
a) 44.1m b)14.2m c) 22.4m d) 42.4m

Answers

Answered by tusharraj77123
9

Answer:

Distance travelled by the marble = 44.1m

Or the option (a) is correct .

Step-by-step explanation:

Given :

Time = 3s

Acceleration due to gravity = 9.8m/s²

Initial velocity = 0 m/s

To find :

The distance travelled by marble while falling down

Formula used :

Using the second equation of motion -:

   \color{red}{ \boxed { \sf {{s = ut +  \dfrac{1}{2} a {t}^{2} }}}}

Where,

s = Distance

u = Initial velocity

t = Time

a = Acceleration due to gravity

Solution :

   :  \implies \sf{s = 0 m/s \times 3 s+  \dfrac{1}{2}  \times 9.8{m/s}^{2} \times  {3}^{2}}  \\  \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    :  \implies \sf{ \dfrac{1}{ \cancel{2}}  \times  \cancel{9.8} {m/s}^{2}\times (3 \times 3)} \\  \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    :  \implies \sf{4.9 \times 9} \\  \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:      :  \implies \sf \fbox{ \color{purple}{44.1m}}

So , the distance travelled by the marble is 44.1m .

Extra information :

First equation of motion -:

\boxed{\sf{u=v+at}}

Where,

u = Initial velocity

v = Final velocity

a = Acceleration

t = Time

Third equation of motion -:

\boxed{\sf{{v}^{2}-{u}^{2}=2as}}

Where,

v = Final velocity

u = Initial velocity

a = Acceleration

s = Distance

Answered by payarelal491
0

Answer:

Marble is dropped from a top of a tower, what distance does it fall after 3 second of dropping. find

its velocity at this point

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