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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Marques Mckay

2 March, 09:11

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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Aryana Guerra · Answer 1

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 dа ta:

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