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uysha [10]
3 years ago
12

At what vertical velocity should no object be launched at in order to achieve a height of 20m?​

Physics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

19.8 m/s

Explanation:

Given:

Maximum vertical displacement of the object (H) = 20 m

Acceleration due to gravity (g) = 9.8 m/s²

At maximum height, the velocity of the object is 0 m/s for a moment. So, final velocity (v) at the maximum height is 0 m/s.

Now, let the initial velocity or velocity at launch be 'u' m/s.

Now, using the following equation of motion for vertical motion:

v^2=u^2-2gH

Rewriting in terms of 'u', we get:

u^2=v^2+2gH\\\\u=\sqrt{v^2+2gH}

Plug in the given values and solve for 'u'. This gives,

u=\sqrt{0+2\times 9.8\times 20}\\\\u=\sqrt{392}\\\\u=19.8\ m/s

Therefore, the vertical velocity at the launch is 19.8 m/s.

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A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside
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Answer:

\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}, level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}

\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt}  = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}

By replacing all known variables:

\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}

The positive sign of the rate of change of the tank level indicates a rising behaviour.

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