Question

A particle is projected with a velocity of
`29.4ms^-^1` . 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

Answer 1

A.88.2m

Answer:

Solution given:

initial velocity[u]=29.4m/s

g=9.8m/s²

maximum range=?

now

we have

`\theta=90°`

maximum range =
`(29.4²*sin90)/(9.8)=88.2m`

Answer 2

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.