Question

A 6 kg bowling ball is dropped from the top of a building. If it hits the ground with a speed of 55.0 m/s, how much PE did it have at the top of the building and how tall was the building​

Answer 1

$answer$

Potential Energy = 9075 J

Height = 151.25 m

Explanation:

Given:

Mass of ball (m) = 6 kg

Velocity (v) = 55 m/s

To Find:

How tall was the building (Height).

Potential Energy

Now, we know that

$\implies{\tt{P.E = mgh}}$

$\implies{\tt{P.E = 6\times 10\times h\;\;[Take\;g=10\;m/s^{2}]}}$

$\implies{\tt{P.E = 60h\;\;...(1)}}$

When it is about to reach the ground,  P.E will fully be converted into K.E.

So,

$\implies{\tt{K.E=P.E...\;(2)}}$

$\implies{\tt{K.E=\dfrac{1}{2}\;mv^{2}}}$

$\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 55^{2}}}$

$\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 3025}}$

$\implies{\tt{K.E=\dfrac{1}{2}\times 18150}}$

$\implies{\boxed{\tt{K.E=9075\;J}}}$

Hence, Potential Energy at the top of the building = 9075 J.

Now, Equalate both the equations,

$\implies{\tt{K.E=P.E}}$

$\implies{\tt{9075=60h}}$

$\implies{\tt{\dfrac{9075}{60}=h}}$

$\implies{\boxed{\tt{h=151.25\;m}}}$

Hence, Height of the building = 151.25 m.

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Answer 2

$\rm \: Given \\ \tt Mass \: of \: ball (m) = 6 kg \\ \tt \: Velocity (v) = 55 m/s \\ \\ \rm To \: Find \\ \tt Height \: of \: the \: building \\ \tt Potential \: Energy \\ \\ \sf \: we \: know \: that \\ {\tt{P.E = mgh}} \\ {\tt{P.E = 6\times 10\times h\;\;[Take\;g=10\;m/s^{2}]}} \\ {\tt{P.E = 60h\;\;...(1)}} \\ \\ \bf \scriptsize{When \: it \: is \: about \: to \: reach \: the \: ground, \:  P.E \: will \: fully \: be \: converted \: into \: K.E.} \\ So, \\ {\tt{K.E=P.E...\;(2)}} \\ {\tt{K.E=\dfrac{1}{2}\;mv^{2}}} \\ {\tt{K.E=\dfrac{1}{2}\times 6\times 55^{2}}} \\ {\tt{K.E=\dfrac{1}{2}\times 6\times 3025}} \\ {\tt{K.E=\dfrac{1}{2}\times 18150}} \\ {\boxed{\tt{K.E=9075\;J}}} \\ \sf\small{ Hence, \: Potential \: Energy \: at \: the \: top \: of \: the \: building = 9075 J.} \\ \\ \bf \: Now, \: Equalate \: both \: the \: equations, \\ \\ {\tt{K.E=P.E}} \\ {\tt{9075=60h}} \\ {\tt{\dfrac{9075}{60}=h}} \\ {\boxed{\tt{h=151.25\;m}}} \\ \rm \: Height \: of \: the \: building = 151.25 m.$