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Lorico [155]
3 years ago
10

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming

air resistance is negligible, find the speed (magnitude of velocity) in meters per second of the shell at launch and 5.4 s after the launch.
Physics
1 answer:
PtichkaEL [24]3 years ago
6 0

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.                    

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

u=v-at

a=-g

u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s

The velocity of the shell at launch and 5.4 s after the launch is given by :

v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

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Answer:68.15m/s

Explanation:

<u><em>Given: </em></u>

v₁=15m/s

a=6.5m/s²

v₁=?

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v₁²=v₁²+2a (x)

<u>Set up:</u>

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Topic Gravitational force amd firld strength.. help me please
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The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

x \approx 6.74\,\mathrm m

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3 years ago
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