Question

A body is projected from aa point such that the horizontal and vertical components of its velocity are
`640ms^-^1` 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

Answer 1

D. 11520m

Answer:

Solution given:

initial velocity[u]=480m/s

g=10m/s²

maximum height=?

now

we have

maximum height=
`(u²sin²\theta)/(2g)`

where

`\theta=90°`

=
`(480²*sin90)/(2*10)=11520m`