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 75 m away.

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

Explanation / Answer

Vo = initial velocity of launch of javelin

Consider the motion along X-direction

Vox = initial velocity = Vo Cos30

t = time

X = distance travelled = 75 m

using the equation

X = Vox t

t = 75/(Vo Cos30) eq-1

consider the motion along the Y-direction

Voy = Vo Sin30

t = time of travel

Yo = initial position = 2 m

Y = final position = 0 m

a = acceleration = - 9.8

using the equation

Y = Yo + Voy t + (0.5) a t2

0 = 2 + (Vo Sin30) t + (0.5) (-9.8) t2

0 = 2 + (Vo Sin30) (75/(Vo Cos30)) + (0.5) (-9.8) (75/(Vo Cos30))2

Vo = 28.5 m/s

consider the motion during throw"

Vi = initial velocity = 0 m/s

Vf = final velocity = 28.5 m/s

a = acceleration = ?

d = distance travelled = 70 cm = 0.70 m

using the equation

Vf2 = Vi2 + 2 a d

28.52 = 02 + 2 a (0.70)

a = 580.2 m/s2

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