Question on: JAMB Physics - 1994
An object is projected with a velocity of 80ms-1 at an angle of 30o to the horizontal.The maximum height reached is
A
20m
B
80m
C
160m
D
320m
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Correct Option: B
max. height = (\frac{U^2\sin^2\theta}{2g})
= (\frac{80 \times 80 \times \sin^2(30)}{2 \times 10})
= 80m
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