Question

A 25-newton weight falls freely from rest from the roof of a building. What is the total distance the weight falls in the first 1.0 second?

Answer 1

Given:-

• Initial velocity ,u = 0m/s

• Time taken,t = 1 s

• Acceleration due to gravity ,g = 9.8m/s²

To Find:-

• Distance covered ,S

Solution:-

Using 2nd Equation of Motion

• s = ut + 1/2at²

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

s is the distance travelled

Substitute the value we get

→ s = 0×1 + 1/2×9.8×1²

→ s = 0 + 1/2 × 9.8×1

→ s = 1/2×9.8

→ s = 4.9 m

Therefore,the distance covered by the object is 4.9 metres .

Answer 2

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A 25-newton weight falls freely from rest from the roof of a building. What is the total distance the weight falls in the first 1.0 second?

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• $Initial \:velocity(u) =0m/s$
• $Time \:taken (t) =1s$
• $Acceleration \:due\: to\: gravity =9.8m/{s}^{2}$

${ { \underbrace{ \mathbb{ \red{To\:PrOve\ }}}}}$

• $Distance\: covered(s)$

${ \color{aqua}{ \underbrace{ \underline{ \color{lime}{ \mathbb{\star SoLuTiOn\star }}}}}}$

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

• $s=distance\: travelled$
• $u=initial\: velocity$
• $t=time\:taken$
• $a=acceleration$

$Substitute \:the\:values$

$s=0\times +\frac{1}{2}\times 9.8\times {1}^{2}$

$s=0 +\frac{1}{2}\times 9.8\times 1$

$s=\frac{1}{2}\times 9.8$

${\boxed {\boxed {s=4.9m}}}$

$\therefore the \:distance\: covered\: by \:the\: object\\ is\: {\boxed {\boxed {4.9m}}}$

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${\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5069}}}}}}}}}}}}}}}$