1.

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

Answer»

R = 16 H
R = 8 H
R = 4 H
R = 2 H

Solution :`H=(U^(2)SIN^(2)45^(@))/(2g)=(U^(2))/(4G)`,
`R=(U^(2)sin90^(@))/(G)=(U^(2))/(g)`
`therefore R=4H`


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