Question

A javelin thrower standing at rest holds the center through a distance of 70 cm as she throws. She angle, not the 45 you might have expected would be able to at 45. In this throw, the javelin hits the ground 72 m away Units aValue Pearsorn

Answer 1

Here, first you need to find the velocity at which the javelin leaves her hand.

Suppose this 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

Now, write an equation for time, using the horizontal motion:
r = v × t
72 = 0.866×V × t
=> t = 83.14 / V

Enter that time in the equation for vertical motion:
s = u×t + a×t²/2
-2 = 0.5×V×83.14/V + (-9.8)×(83.14/V)²/2
-2 = 41.57 - 33870.07 / V^2

=> 33870.07 / V^2 = 43.57

=> V^2 = 33870.07 / 43.57

=> V = 27.88 m/s

Now find the acceleration of the javelin while throwing:

v² = u² + 2×a×s
27.88² = 0² + 2×a×0.7
=> a = (27.88²) / (2×0.7) =  555.3 m/s^2 (Answer)

Answer in two significant digits = 560.0 m/s^2