A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of 70 cm as she throws. She releases the javelin 2.0 m above the ground traveling at an angle of 30 ∘ above the horizontal. Top-rated javelin throwers do throw at about a 30 ∘ angle, not the 45 ∘ you might have expected, because the biomechanics of the arm allow them to throw the javelin much faster at 30 ∘ than they would be able to at 45 ∘ . In this throw, the javelin hits the ground 50 m away.

What was the acceleration of the javelin during the throw? Assume that it has a constant acceleration.

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banjosimage banjosimage · Answer 1 · 2020-02-19T01:07:40+01:00

Answer:

The acceleration of the javelin during the throw, assume that it has a constant acceleration is: 377.92 m/s2

Explanation:

The first step is to get the velocity at which the javelin leaves the hand value (V):

V horizontal component:

Vh = V×cos(30) = 0.866×V

V vertical component:

Vv = V×sin(30) = 0.5×V

Using horizontal motion data:

r = v × t

50 = 0.866×V × t

t = 57.73/V

Use time gotten in the vertical motion equation:

s = u×t + a×t²/2

-2 = 0.5×V×57.73/V + (-9.8)×(57.73/V)²/2

V = 23.002 m/s

Now let´s get the acceleration of the javelin while throwing:

v² = u² + 2×a×s

23.002² = 0² + 2×a×0.7

a = 377.92 m/s²