3 years ago

Can you help me on this fast just the second one

1 answer:

3 0

4. E
5. D
6. F

Hope this helps

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julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be $\frac{1876.8}{g}$

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

$v_{y}=u_{y}+(-g) t$

$u_{y}= u\sin \phi$

and $v_{y}=0$

therefore $\frac{u\sin \phi }{g} =t$

writing equation of motion in horizontal direction:

$s_{x}=u_{x}t$

$u_{x} = u\cos \phi$

therefore the equation becomes $s_{x}=\frac{u^{2} \sin \phi \cos \phi}{g}$

therefore horizontal distance traveled = $\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}$

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