Question

A ball of mass 0.5 kg is thrown up with a velocity of 15m/s. Find its potential energy at the highest point

Answer 1

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HERE'S THE ANSWER...

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♠️ We know the velocity at max height is 0

⏺️ Initial velocity ( u ) = 15 m/s

⏺️Final Velocity ( v ) = 0

⏺️ Acceleration( a ) = - g = 10 m/s^2

⏺️ Max Height ( s ) = ?

▶️ Using third equation of motion

✔️ $\texttt{ ( v )^{2} - ( u )^{2} \:=\: 2as}$

=> $\texttt{ ( 0 )^{2} - ( 15 )^{2} \:=\: 2 (- 9.8 )s}$

=> $\texttt{ - 225 \:=\: 2 (- 10 ) s}$

=> $\texttt{ - 225 \:=\: -20 s}$

=> $\texttt{ s \:=\: 225/20}$

=> $\texttt{ s \:=\: 11.25\: meter ( m )}$

♠️ Now For potential energy ( U )

✔️ $\texttt{ U\:=\: mgh}$

⏺️mass ( m ) = 0.5 kg

⏺️Acc. due to gravity ( g ) = 10 m/s^2

⏺️ Height ( s = h ) = 11.25 m

=> $\texttt{ U\:=\: mgh}$

=> $\texttt{ U\:=\: ( 0.5 )( 11.25)( 10)}$

=> $\texttt{ U\:=\: ( 5 )( 11.25)}$

=> $\boxed{ U\:=\: 56.25 Joules( J )}$

♠️ Potential energy at highest point will be ( 56.25 J )

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Answer 2

v^2 - u^2 = 2(-g) h

0 - 15^2 = 2 × -10 × h

-225 / -20 = h

height = 11.25 m

now pot energy ( PE )

PE = m×g×h

= 0.5 × 10 × 11.25

PE = 56.25 J