A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of 70

Question

3 years ago

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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.

Physics

1 answer:

JulijaS [17] · Answer 1 · 2022-07-17T09:10:06+03:00

Answer:

a = 580 m/s^2

Explanation:

Given:

- Distance for accelerated throw s_a = 70 cm

- Angle of throw Q = 30 degrees

- Distance traveled by the javelin in horizontal direction x(f) = 75 m

- Initial height of throw y(0) = 0

- Final height of the javelin y(f) = -2 m

Find:

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

Solution:

- Compute initial components of the velocity:

V_x,i = V*cos(30)

V_y,i = V*sin(30)

- Use second equation of motion in horizontal direction:

x(f) = x(0) + V*cos(30)*t

75 = 0 + V*cos(30)*t

t = 75 /V*cos(30)

- Use equation of motion in vertical direction:

y(f) = y(0) + V_y,i*t + 0.5*g*t^2

Subs the values:

-2 = 0 + V*sin(30)*75/V*cos(30) - 4.905*(75/Vcos(30))^2

-2 = 75*tan(30) - 4.905*(5625/V^2*cos^2(30))

V^2 = 4.905*5625 / (2 + 75*tan(30))*cos^2(30)

V^2 = 812.0633

V = 28.5 m/s

- Use the third equation of motion in the interval of the throw:

V^2 = U^2 + 2*a*s_a

28.5^2 = 2*a*0.7

a = 580 m/s^2