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2 years ago

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If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f

ast must the fish spit the water to hit its target? The insect is at the highest point of the trajectory of the spit water. Use g = 10 m/s2.

1 answer:

Burka [1]2 years ago

3 0

Answer:

The speed is $v = 9.8 \ m/s$

Explanation:

From the question w are told that

The angle made is $\theta = 30^o$

The distance above the surface of the water is $h_{max} = 1.2 \ m$

The value of $g = 10 \ m/s^2$

The maximum height attained by the fish is mathematically evaluate as

$h_{max} = \frac{v^2 sin ^2 \theta }{2g }$

Making v which is the speed of the fish the subject of the formula

$v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }$

substituting values

$v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }$

$v = 9.8 \ m/s$

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The potential energy of the block is given by:
V = m*g*h
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2 years ago

a rocket initially at rest on the ground lifts off vertically with a constant acceleration of 2.0 x 10^1 meters per second^2. Ho

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Here's the formula for the distance covered by an accelerating body in some amount of time ' T '. This formula is incredibly simple but incredibly useful. It pops up so often in Physics that you really should memorize it:

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An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal

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Answer:

The value is $T_t = 2.5659 \ s$

Explanation:

From the we are told that

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$0^2 = 8^2 + 2 * - 9.8 * H$

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So the initial height is mathematically represented as

$s = h - H$

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