ZJSG09112007
contestada

A body is projected from aa point such that the horizontal and vertical components of its velocity are [tex]640ms^-^1[/tex] and [tex]480 ms^-^1[/tex] respectively . (Take g = [tex]10ms^-^1[/tex] )
i. Calculate the greatest height attained above the point of projection
A.600m B.1215 m C.1521 m D. 11520m E. 20480m

Respuesta :

msm555

D. 11520m

Answer:

Solution given:

initial velocity[u]=480m/s

g=10m/s²

maximum height=?

now

we have

maximum height=[tex]\frac{u²sin²\theta}{2g}[/tex]

where

[tex]\theta=90°[/tex]

=[tex]\frac{480²*sin90}{2*10}=11520m[/tex]

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