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anonymous

  • 5 years ago

HELP!! On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-fall acceleration on the moon is 1/6 of its value on earth. Suppose he hit the ball with a speed of 35 m/s at an angle 25 degree above the horizontal. How long was the ball in flight? How far did it travel? Ignoring air resistance, how much farther would it travel on the moon than on earth?

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  1. nowhereman
    • 5 years ago
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    Assume earth and moon as flat and the gravity always going downwards. Then you can look at vertical and horizontal speed / position separately.

  2. anonymous
    • 5 years ago
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    which formula do i use or how to approach this? Xf = Xi + (Vx)i t ?

  3. nowhereman
    • 5 years ago
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    I don't know those variables. But if you have constant acceleration called a, time t, name speed v and position x, then you get \[a(t) = \dot v(t)\quad v(t) = \dot x(t)\] so \[a(t) = \ddot x(t) ⇒ x(t) = x(0) + \int_0^t{v(0) + \int_0^{s_2}{a(s_1)ds_1}ds_2} = x(0) + \int_0^t{v(0) + a\cdot s_1ds_1}\] \[=x(0) + v(0)t + \frac a 2 t^2\]

  4. anonymous
    • 5 years ago
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    hmm thanks but i dont have the "t" so do i find that first

  5. nowhereman
    • 5 years ago
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    That is the correct approach. Find t so that the vertical component is zero again, then find the horizontal component for that t.

  6. nowhereman
    • 5 years ago
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    And remember the above quadratic formula is for a = 9.81 N/kg in vertical direction on the earth and for a = 0 in horizontal direction.

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