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 72 m away.

Explanation / Answer

Let throwing speed be v.

vertical speed = v sin 30 degree = 0.5v

horizontal speed = v cos 30 degree = 0.866v

time taken t= 72/0.866v

now, writing second equation of motion in vertical,

h= u t + 0.5gt^2

2 = -0.5v * 72/0.866v + 4.9 (72/0.866v)^2

2 = -41.57 + 4.9* (72/0.866v)^2

43.57/4.9 = (72/0.866v)^2

(72/0.866v) = sqrt(43.57/4.9) = 2.982

v = (72/0.866)/2.982

= 27.88 m/s

Now acceleration can be found by using third equation of motion,

v^2 = 0+2as

a = v^2/2s = 27.88^2/(2*0.70)

= 555 m/s^2 answer

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