For an object thrown at `45^(@)` to the horizontal, the maximum height H and horizontal range R are related as

Options

A. R = 16 H
B. R = 8 H
C. R = 4 H
D. R = 2 H

Answer

C

Explanation

`H=(U^(2)sin^(2)45^(@))/(2g)=(U^(2))/(4g)`,
`R=(U^(2)sin90^(@))/(g)=(U^(2))/(g)`
`therefore R=4H`