ZJSG09112007
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A particle is projected with a velocity of [tex]29.4ms^-^1[/tex] . Find it's maximum range on a horizontal plane through the point of projection.
A.88.2m B.44.1m C.32.6m D.29.4m E.14.7m

Respuesta :

msm555

A.88.2m

Answer:

Solution given:

initial velocity[u]=29.4m/s

g=9.8m/s²

maximum range=?

now

we have

[tex]\theta=90°[/tex]

maximum range =[tex]\frac{29.4²*sin90}{9.8}=88.2m[/tex]

ItzTds

The initial velocity is,

→ u = 29.4 m/s

General assumption,

→ g = 9.8m/s²

→ θ = 90°

Then the maximum range is,

→ (29.4² × sin90)/9.8

→ 88.2 m

Hence, option (A) is answer.

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