Home » Classroom » JAMB Physics 1994 Question

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

Please share this, thanks:

#JAMB #JAMB

Add your answer

Notice: Please post responsibly.

No responses