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a golfer can hit a golf ball a horizontal distance of over 300 m on a good drive. what maximum height would a 301.5 m drive reach if it were launched at an angle of 25.0 degrees to the ground?

a. 140.6 m
b. 70.3 m
c. 150.8 m
d. 332 m

huh??!?

2007-12-16 16:15:45 · 3 answers · asked by Anonymous in Science & Mathematics Physics

3 answers

Range and height are related by the equation
4H = R tan θ = 301.5 tan 25 = 386
H = 35.15m
--------------------------
None of the answer is correct.

2007-12-16 21:08:08 · answer #1 · answered by Pearlsawme 7 · 0 0

b. 70.3 m

Here is my quick, non-computational, conceptual approach to see if any answer stands out as likely. A launched object follows a parabolic path, which is bilaterally symmetrical. A triangle inscribed within that parabola will be isosceles and have angles at the points of launch and landing equal to 25.0 degrees and a base equal to 301.5m. Bisecting that triangle with a vertical line will create two right triangles with angles of 25, 90, and 65 degrees. The line between the 25 and 90 degree angles, the base of each new right triangle, will be 150.525 m long. The maximum height of the arc of travel will be equal to the length of the vertical, bisecting line, a line which will be significantly shorter than the 150.525 m line (as a quick sketch of such a triangle will show). The only answer that gives a measure significantly less than 150.525 m is (b). Assuming that one of the four choices provided is correct, I would choose (b).

* Please note that I have modified this from my original answer, primarily by correcting the name for the type of triangle inscribed inside the parabola. I originally, hastily, wrote equilateral and have corrected that to isosceles.

2007-12-17 00:44:16 · answer #2 · answered by Anonymous · 0 0

Horizontal:

x - xo = vox t

x - xo = 301.5 m
vox = vo cos 25.0 degrees

Solve for t in terms of vo.

t = 301.5 m / ( vo cos 25 )

Vertical:

y - yo = voy t - 1/2 g t²

y - yo = 0 ( the ball is back on the ground again )
voy = vo sin 25
g = 9.8 m/s²

Plug in your expression for t. You now have an equation with only one variable, vo. Solve for vo.

Now that we have the initial velocity, we can get the height:

2 g h = ( vo sin 25 )²

Plug in g and vo and solve for h.

2007-12-17 00:36:11 · answer #3 · answered by jgoulden 7 · 0 0

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