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Jobisdone [24]
3 years ago
14

A ball is thrown from 1 m above the ground. The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal. What

is the maximum height of the ball above the ground
Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

9.4 m

Explanation:

This is a projectile motion. The maximum height above the level projection is given by

H = \dfrac{(v_0\sin\theta)^2}{2g}

v_0 is the initial velocity, \theta is the angle of projection above the horizontal and g is the acceleration of gravity.

H = \dfrac{(20\sin40^\circ)^2}{2\times9.8} = \dfrac{165.27}{19.6} = 8.4\text{ m}

The maximujm height above the ground is 1 m + 8.4 m = 9.4 m

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θ = 66.90°

Explanation:

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Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 2
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\frac{500}{10 ^{ - 4} }

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<u>Elephant</u>

<u>Area</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>0</u><u>cm</u><u>²</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>x</u><u> </u><u>1</u><u>0</u><u>^</u><u>(</u><u>-</u><u>2</u><u>)</u><u>b</u><u> </u><u>m</u><u>²</u>

<u>Force</u><u> </u><u>=</u><u> </u><u>mg</u><u> </u><u>=</u><u> </u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>N</u>

<u>Pressure</u><u> </u><u>=</u><u> </u>

<u>\frac{40000}{2.5 \times  {10}^{ - 2} }</u>

<u>= 1.6 \times  {10}^{6}</u>

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A wave has a frequency of 440 hz and a wavelength of 0.77m. What is the velocity of the wave
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