Question

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A body is thrown upward reaches to a height of 100m in 10 sec. Find the initial velocity. ( g = 10m/s ² )​

Answer 1

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Q) A body is thrown upwards reaches to a height of 100m in 10 sec . Find the initial velocity .

( g = 10m/s )

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${\large{\rm{\underline{\star\;Given:-}}}}$

• Height = 100 m
• Time = 10 sec
• g = 10 m/s²

${\large{\rm{\underline{\star\;To\;Find:-}}}}$

• The initial velocity = ?

${\large{\rm{\underline{\star\; Solution:-}}}}$

In this case ,

• The height would be the distance s .
• The "g" would act as acceleration .
• g would have a negative sign as it is acting opposite to the direction .

______

• s = 100 m
• t = 10 s
• a = -g = -10m/s²
• u (initial velocity) = ?

Using the second Equation of Motion -

$\boxed{ \sf{ s = ut + \frac{1}{2}a {t}^{2} }}$

put the values ,

$\mapsto \rm \: 100 = u \times 10 + \frac{1}{ \cancel2} ( - \cancel{10}) {(10)}^{2}$

$\mapsto \rm \: 100 = 10u + ( - 5)(100)$

$\mapsto \rm \: 100 =10 u - 500$

$\mapsto \rm \: 100 + 500 = 10u$

$\mapsto \rm \: \cancel{10}u = \cancel{600}$

$\mapsto \underline{ \boxed{ \rm{ \pink{u = 60 \: \frac{m}{s} }}}}$

So ,

The initial velocity would be 60 m/s .

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★ Know More :-

Equations of motion :

1. $\tt \: v = u + at$
2. $\tt \: s = ut + \frac{1}{2} a {t}^{2}$
3. $\tt \: {v}^{2} - {u}^{2} = 2as$

Here ,

• u = initial velocity.

• v = final velocity.

• a = acceleration.

• s = distance .

• t = time.

Answer 2

Maximum height , S = 100 m

Time taken = 10 sec

Initial velocity = ?

g = - 10 m/s² [ Acting against the initial direction of motion of the body.]

Using the second Equation of Motion

s = ut + at²/2

100 = u×10 + 1/2 × (-10) × (10)²

100 = 10u + (−5)(100)

100 = 10u − 500

10u = 100 + 500

10u = 600

u = 60 m/s