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vivado [14]
3 years ago
5

player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the hor

izontal velocity component of the ball to the nearest tenth of a m/s?
Physics
2 answers:
seropon [69]3 years ago
7 0

Answer

22.5 m/s

Explanation

We shall use the trigonometric ratio cosine to find the horizontal component.

cos = adjacent/hypotenuse

Adjacent is the horizontal and hypotenuse is the fly speed.

cos 30° = horizontal / 26

horizontal velocity = 26 × cos 30°

                                = 26 × 0.866

                                = 22.5166

                                 = 22.5 m/s

Ad libitum [116K]3 years ago
4 0

Solution

In this Question we have given,

Velocity, V=26\frac{m}{s}

angle,α=30 degrees

We have to find horizontal component of velocity,

V_{x}=V\times Cos\alpha

V_{x}=26\times Cos\30

V_{x}=26\times 0.866

V_{x}=22.51\frac{m}{s}

V_{x}=23\frac{m}{s}

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Explanation:

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The maximum speed v given as

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0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

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A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

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koban [17]

Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

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With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

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The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

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The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

p_f = p_{fB}+p_{fS}

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b) -32 kg m/s

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c) 64 N

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d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

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