A particle is projected from level ground its kineitc energy K changes due to gravity so that `K_("max")//K_("min")=9`. The ratio of the range to the

Question

A particle is projected from level ground its kineitc energy K changes due to gravity so that `K_("max")//K_("min")=9`. The ratio of the range to the

A particle is projected from level ground its kineitc energy K changes due to gravity so that `K_("max")//K_("min")=9`. The ratio of the range to the maxir, un height attained during its fight is
A. `sqrt(2)`
B. `4sqrt(2)`
C. 1.5
D. none

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`KE_("max")=(1)/(2)mv^(2)` (at the point of projection).
`KE_("min")=(1)/(2) mv^(2)cos ^(2)theta`
`(v^(2))/(v^(2)cos^(2)theta)=9 rArr cos theta=(1)/(3)`
as `R=(v^(2)sin2 theta)/(g), H=(v^(2)sin^(2) theta)/(2g)`
`R//H=4 cot theta`

5 views Answered 2023-02-05 15:29