An object is projected with a velocity of 80ms-... - JAMB Physics 1994 Question
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
correct option: b
max. height = \(\frac{U^2\sin^2\theta}{2g}\)
= \(\frac{80 \times 80 \times \sin^2(30)}{2 \times 10}\)
= 80m
= \(\frac{80 \times 80 \times \sin^2(30)}{2 \times 10}\)
= 80m
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