Question

A golfer hits a 42g ball , which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s Determine the height the ball will rise after the bounce .show all your work

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Answer 1

$\fbox\red{A}\fbox\red{n}\fbox\red{s}\fbox\red{w}\fbox\red{e}\fbox\red{r}$

Explanation:

Given:

v₀ = 15.6 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy

Δy = 12.2 m

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\LARGE{ \boxed{ \rm{ \green{Answer:}}}}

\LARGE{ \boxed{ \rm{ \green{Answer:}}}}

Given,

•

The initial speed is 15.6 m/s

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The mass of the ball is 42g = 0.042kg

Finding the initial kinetic energy,

\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}

\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}

⇛ KE = (1/2)mv²

⇛ KE = (1/2)(0.042)(15.6)²

Answer 2

ᎪNᏚᏔᎬᎡ

The height of the ball will rise the bounce = 12.41m.

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Explanation :

We are given with the mass, initial velocity, final velocity, and acceleration due to gravity of a ball, that is,

Mass of ball, m = 42 g.

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Initial velocity of ball, u = 15.6 m/s.

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Final velocity of ball, v = 0 m/s.

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

We have to find out the height of the ball will rise the bounce.

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We know that, if we are given with the mass of a ball, initial velocity of ball, final velocity of ball, and acceleration due to gravity, we know the third equation of motion, that is,

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$\rm{\hookrightarrow v^2 = u^2 + 2gh}$

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Substituting the given values in the formula :

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$\tt{:\implies 0^2 = 15.6^2 + 2 \times (-9.8) \times h}$

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$\tt{:\implies 0 = 243.36 + 2 \times (-9.8) \times h}$

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$\tt{:\implies 0 = 243.36 + ( - 19.6) \times h}$

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$\tt{:\implies 0 = 243.36 - 19.6 \times h}$

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$\tt{:\implies -243.36 = - 19.6 \times h}$

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$\tt{:\implies -243.36 = - 19.6 \times h}$

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$\tt{:\implies -243.36 = -19.6h}$

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$\tt{:\implies h = \dfrac{ \cancel- 243.36}{ \cancel- 19.6} }$

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$\tt{:\implies h = \cancel{\dfrac{243.36}{19.6} }}$

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$\tt{:\implies \boxed{ \frak{ \pink{h = 12.41 \: m}}}}$

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Hence, the height of the ball will rise the bounce is 12.41 m.

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