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myrzilka [38]
4 years ago
12

A 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. How far did t

he ball travel before it hit the ground? Round the answer to the nearest meter.
Physics
2 answers:
Furkat [3]4 years ago
3 0
What are your answers?

Agata [3.3K]4 years ago
3 0

Answer:

60m (to nearest meter)

Explanation: How far the ball traveled refers to the range covered by the ball before it hits the ground. For an object projected at an angle of ∝ to the horizontal, the range, R covered is given by;

R=u^{2}sin2\alpha/2g

Where u is the initial velocity and g is acceleration due to gravity.

u=26mls\\\alpha=30^{o}\\g=9.81m/s^{2}

The value of g could be safely assumed to be 9.8m/s^{2}

Therefore;

R=26^{2}sin2(30)/9.8

R=(26*26*sin60)/9.8

R=676*0.8660/9.8

R=59.68m which is approximately 60m to nearest meter.

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On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

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3\lambda/2 = L

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Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

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