JAMB - Physics (1994 - No. 3)
An object is projected with a velocity of 80ms-1 at an angle of 30o to the horizontal.The maximum height reached is
20m
80m
160m
320m
Giải thích
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