1.

A projectile is projected in vacuum at an angle 0, then square of the time it takes to reach the highest point shall be :

Answer»

2g TIMES the GREATEST height
G times the greatest height
g/2 times the greatest height
2/g times the greatest height

Solution :Here `H=(u^(2)sin^(2)THETA)/(2g)` and time to reach the highest point
`T=(usintheta)/g`or `T^(2)=(u^(2)sin^(2)theta)/g^(2)`
`T^(2)/H=(u^(2)sin^(2)theta)/g^(2)XX(2g)/(u^(2)sin^(2)theta)=2/g`
`T^(2)=2/g.H`


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