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elena55 [62]
2 years ago
5

A boy throws rocks with an initial velocity of 12m/s [down] from a 20 m bridge into a river. Consider the river to be at a heigh

t of 0 m. Draw a velocity-time and position-time graph. Use up as position and ignore air resistance.
Physics
1 answer:
stira [4]2 years ago
4 0

Answer:

Please find attached the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel

Explanation:

The given parameters are;

The initial velocity with which the boy throws the rock, u = 12 m/s

The direction in which he throws the rock = Down

The height from which the rock was thrown, h = 20 m

The height at which the river is located = 0 m

The kinematic equation of motion, of the rock can be given as follows;

h = u·t + 1/2·g·t²

Where;

t = The time of motion of the rock

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the rock is thrown = 20 m

Substituting the known values into given equation, we get;

20 = 12·t + 1/2·9.8·t² = 12·t + 4.9·t²

4.9·t² + 12·t - 20 = 0

t = (-12 ± √(12² - 4×4.9×(-20)))/(2 × 4.9) = (-12 ± √(242))/(9.8)

t ≈ -3.587 seconds or t ≈ 1.138 seconds

Attached please find the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel.

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Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

T\ cos\theta-mg=0

T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

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On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Hence, this is the required solution.

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3 years ago
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