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## In a football game a kicker attempts a field goal. The ball remains in contact with the kicker’s foot for 0.0580 s, during which time it exp

Question

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker’s foot for 0.0580 s, during which time it experiences an acceleration of 376 m/s2. The ball is launched at an angle of 59.9° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.

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Physics
3 years
2021-08-29T05:47:24+00:00
2021-08-29T05:47:24+00:00 1 Answers
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## Answers ( )

Answer:V₀ₓ = 10.94 m/s

V₀y = 18.87 m/s

Explanation:To find the launch velocity, we use 1st equation of motion.

Vf = Vi + atwhere,

Vf = Final Velocity of Ball = Launch Speed = V₀ = ?

Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)

a = acceleration = 376 m/s²

t = time = 0.058 s

Therefore,

V₀ = 0 m/s + (376 m/s²)(0.058 s)V₀ = 21.81 m/sNow, for x-component:

V₀ₓ = V₀ Cos θwhere,

V₀ₓ = x-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀ₓ = (21.81 m/s)(Cos 59.9°)V₀ₓ = 10.94 m/sfor y-component:

V₀ₓ = V₀ Sin θwhere,

V₀y = y-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀y = (21.81 m/s)(Sin 59.9°)V₀y = 18.87 m/s