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BartSMP [9]
3 years ago
5

A body is projected from aa point such that the horizontal and vertical components of its velocity are

n.net/?f=640ms%5E-%5E1" id="TexFormula1" title="640ms^-^1" alt="640ms^-^1" align="absmiddle" class="latex-formula"> and 480 ms^-^1 respectively . (Take g = 10ms^-^1 )
i. Calculate the greatest height attained above the point of projection
A.600m B.1215 m C.1521 m D. 11520m E. 20480m
Mathematics
1 answer:
IRINA_888 [86]3 years ago
3 0

D. 11520m

Answer:

Solution given:

initial velocity[u]=480m/s

g=10m/s²

maximum height=?

now

we have

maximum height=\frac{u²sin²\theta}{2g}

where

\theta=90°

=\frac{480²*sin90}{2*10}=11520m

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