A javelin thrower standing at rest holds the center of the javelin behind her he

Question

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 62 m away. What was the acceleration of the javelin during the throw?

Assume that it has a constant acceleration.

Explanation / Answer

velocity is V.

Horizontal component of V:
Vh = V×cos(30) = 0.866×V

Vertical component of V:
Vv = V×sin(30) = 0.5×V

Write an equation for time, using the horizontal motion:
r = v × t
62 = 0.866×V × t
t = 71.59/V

Enter that time in the equation for vertical motion:
s = u×t + a×t²/2
-2 = 0.5×V×71.59/V + (-9.8)×(71.59/V)²/2
V = 25.78 m/s

Now find the acceleration of the javelin while throwing:

v² = u² + 2×a×s
25.78² = 0² + 2×a×0.7
a = 474.5 m/s²

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