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A golf ball is hit at a velocity of 42.0 m/s at an angle of 42.2° on Earth and on the Moon where the acceleration due to gravity is 1.62 m/s². How much higher will the maximum height be on the Moon compared to the maximum height on Earth?

Sagot :

jacob193

Answer:

Approximately [tex]205\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] on Earth and that air resistance is negligible.)

Explanation:

Initial vertical velocity of the golf ball:

[tex]\begin{aligned} & (\text{initial vertical velocity}) \\ =\; & (\text{initial velocity})\, \sin(\text{angle of elevation}) \\ =\; & (42.0\; {\rm m\cdot s^{-1}}})\, \sin(42.2^{\circ}) \\ \approx\; & 28.212\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Let [tex]u_{y}[/tex] denote the initial vertical of the ball. When the ball is at maximum height, the vertical velocity [tex]v_{y}[/tex] of the ball will be [tex]0[/tex]. Let [tex]x_{y}[/tex] denote the vertical displacement of the ball (height of the ball.)

Let [tex]a_{y}[/tex] denote the vertical acceleration of the ball. Under the assumptions, the vertical acceleration of the ball during the flight will be constantly [tex](-g)[/tex].

The SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] relates these quantities. Rearrange this equation to find the maximum vertical displacement of the ball (value of [tex]x_{y}[/tex] when [tex]v_{y} = 0\; {\rm m\cdot s^{-1}}[/tex].)

[tex]{v_{y}}^{2} - {u_{y}^{2} = 2\, a_{y}\, x_{y}[/tex].

[tex]\begin{aligned}x_{y} &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a_{y}} \end{aligned}[/tex].

On the Earth, [tex]a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. Therefore:

[tex]\begin{aligned}x_{y} &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a_{y}} \\ &\approx \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (28.212\; {\rm m\cdot s^{-1}})^{2}}{2\, ((-9.81)\; {\rm m\cdot s^{-2}})} \\ &\approx 40.57\; {\rm m}\end{aligned}[/tex].

On the Moon, it is given that [tex]g = 1.62\; {\rm m\cdot s^{-2}}[/tex], such that [tex]a_{y} = (-g) = (-1.62)\; {\rm m\cdot s^{-2}}[/tex]. Therefore:

[tex]\begin{aligned}x_{y} &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a_{y}} \\ &\approx \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (28.212\; {\rm m\cdot s^{-1}})^{2}}{2\, ((-1.62)\; {\rm m\cdot s^{-2}})} \\ &\approx 245.65\; {\rm m}\end{aligned}[/tex].

The difference between the maximum heights will be approximately:

[tex](245.65\; {\rm m}) - (40.57\; {\rm m}) \approx 205\; {\rm m}[/tex].

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