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 About the author Aubrey
[tex] \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.[/tex] Reply
[tex]answer[/tex] 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 [tex]\implies{\tt{P.E = mgh}}[/tex] [tex]\implies{\tt{P.E = 6\times 10\times h\;\;[Take\;g=10\;m/s^{2}]}}[/tex] [tex]\implies{\tt{P.E = 60h\;\;…(1)}}[/tex] When it is about to reach the ground, P.E will fully be converted into K.E. So, [tex]\implies{\tt{K.E=P.E…\;(2)}}[/tex] [tex]\implies{\tt{K.E=\dfrac{1}{2}\;mv^{2}}}[/tex] [tex]\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 55^{2}}}[/tex] [tex]\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 3025}}[/tex] [tex]\implies{\tt{K.E=\dfrac{1}{2}\times 18150}}[/tex] [tex]\implies{\boxed{\tt{K.E=9075\;J}}}[/tex] Hence, Potential Energy at the top of the building = 9075 J. Now, Equalate both the equations, [tex]\implies{\tt{K.E=P.E}}[/tex] [tex]\implies{\tt{9075=60h}}[/tex] [tex]\implies{\tt{\dfrac{9075}{60}=h}}[/tex] [tex]\implies{\boxed{\tt{h=151.25\;m}}}[/tex] Hence, Height of the building = 151.25 m. ∴ For more Information visit: https://brainly.in/question/33914988 Reply
[tex] \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.[/tex]
[tex]answer[/tex]
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
[tex]\implies{\tt{P.E = mgh}}[/tex]
[tex]\implies{\tt{P.E = 6\times 10\times h\;\;[Take\;g=10\;m/s^{2}]}}[/tex]
[tex]\implies{\tt{P.E = 60h\;\;…(1)}}[/tex]
When it is about to reach the ground, P.E will fully be converted into K.E.
So,
[tex]\implies{\tt{K.E=P.E…\;(2)}}[/tex]
[tex]\implies{\tt{K.E=\dfrac{1}{2}\;mv^{2}}}[/tex]
[tex]\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 55^{2}}}[/tex]
[tex]\implies{\tt{K.E=\dfrac{1}{2}\times 6\times 3025}}[/tex]
[tex]\implies{\tt{K.E=\dfrac{1}{2}\times 18150}}[/tex]
[tex]\implies{\boxed{\tt{K.E=9075\;J}}}[/tex]
Hence, Potential Energy at the top of the building = 9075 J.
Now, Equalate both the equations,
[tex]\implies{\tt{K.E=P.E}}[/tex]
[tex]\implies{\tt{9075=60h}}[/tex]
[tex]\implies{\tt{\dfrac{9075}{60}=h}}[/tex]
[tex]\implies{\boxed{\tt{h=151.25\;m}}}[/tex]
Hence, Height of the building = 151.25 m.
∴ For more Information visit:
https://brainly.in/question/33914988