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A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler’s hand?

The aim of this question is to understand how to implement and apply kinematic equations of motion.

Kinematics is the branch of physics dealing with objects in motion. Whenever a body moves in a straight line, then the equations of motion can be described by the following formulae:

\[ v_{ f } \ = \ v_{ i } + a t \]

\[ S = v_{i} t + \dfrac{ 1 }{ 2 } a t^2 \]

\[ v_{ f }^2 \ = \ v_{ i }^2 + 2 a S \]

For the vertical upward motion:

\[ v_{ f } \ = \ 0, \ and \ a \ = \ -9.8 \]

In case of vertical downward motion:

\[ v_{ i } \ = \ 0, \ and \ a \ = \ 9.8 \]

Where $ v_{ f } $ and $ v_{ i } $ are the final and initial speed, $ S $ is the distance covered, and $ a $ is the acceleration.

Expert Answer

The given motion can be divided into two parts, vertically upward motion and vertically downward motion.

For the vertically upward motion:

\[ v_i \ = \ 8.20 \ m/s \]

\[ v_f \ = \ 0 \ m/s \]

\[ a \ = \ -g \ = \ 9.8 \ m/s^{ 2 } \]

From the first equation of motion:

\[ v_{ f } \ = \ v_{ i } + a t \]

\[ \Rightarrow t \ = \ \dfrac{ v_{ f } \ –  v_{ i } }{ a } … \ … \ … \ ( 1 ) \]

Substituting values:

\[ t \ = \ \dfrac{ 0 \ –  20 }{ -9.8 } \]

\[ \Rightarrow t \ = \ \dfrac{ -20 }{ -9.8 } \]

\[ \Rightarrow t \ = \ 2.04 \ s \]

Since the body has the same acceleration and has to cover the same distance during the vertically downward motion, it will elapse the same amount of time as the vertically upward motion. So:

\[ t_{ total } \ = \ 2 \times t \ = \ 4.08 \ s \]

Numerical Results

\[ t_{ total } \ = \ 4.08 \ s \]

Example

Calculate the distance covered by the bowling pin during the upward motion.

For the vertically upward motion:

\[ v_i \ = \ 8.20 \ m/s \]

\[ v_f \ = \ 0 \ m/s \]

\[ a \ = \ -g \ = \ 9.8 \ m/s^{ 2 } \]

From the 3rd equation of motion:

\[ v_{ f }^2 \ = \ v_{ i }^2 + 2 a S \]

\[ \Rightarrow S \ = \ \dfrac{ v_{ f }^2 \ – \ v_{ i }^2 }{ 2 a } \]

Substituting values:

\[ \Rightarrow S \ = \ \dfrac{ ( 0 )^2 \ – \ ( 8.20 )^2 }{ 2 ( -9.8 ) } \]

\[ \Rightarrow S \ = \ \dfrac{ – 67.24 }{ – 19.6 } \]

\[ \Rightarrow S \ = \ 3.43 \ m \]

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