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

Answer 1

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