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 degree above the horizontal. Top-rated javelin throwers do throw at about a 30 angle, not the 45 degree you might have expected, because the biomechanics of the arm allow them to throw the javelin much faster at 30 degree than they would be able to at 45 degree. In this throw, the javelin hits the ground 66 m away. What was the acceleration of the javelin during the throw? Assume that it has a constant acceleration. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Assuming that the javelin has a horizontal velocity of vx ( = v cos30 = v/2 )as it travels in air.

Time taken to hit ground = 132/v

During this time, vertical distance travelled = -2m ( assuming upward direction as +ve )

Using equation:

S = ut + 0.5at2

Here, u = vy = v sin 30 = 0.86 v , a = -9.8 m/s2 , t = 132/v , S = -2,

Putting values and solving we get:

v = 27.18 m/s

Assuming acceleration of javelin as 'a' we have:

v2-u2 = 2aS

Here, u = 0, S = 0.7 m

Putting values, we get:

a = v2/2S = 27.182/1.4 = 527.9 m/s2

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