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Keeping the velocity of projection constant, the angle of projection is increased from 0° to 90º. Then the maximum height of the projectile
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a
goes on increasing up to 900
b
decreases up to 900
c
increase up to 450 and decreases beyond it
d
decreases up to 450 and increases beyond it
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Correct option is A
Hmax = u2sin2α / 2gAt α = 00 , Hmax = 0 and α = 900 , Hmax has the maximum value of u2/2g , So option (1) is right.Similar Questions
A ball is thrown with a velocity of u making an angle with the horizontal. Its velocity vector normal to initial vector (u) after a time interval of
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