Question

Prove that a) is dimensionally correct

Answer 1

is this the whole question ? :o

Answer 2

a) \[\Delta x = \frac{vi ^{2}\sin2 \theta }{ g }\]

Answer 3

Just rewrite the equation using the appropriate dimensions...noting that sine(2Theta) is a scalar value:
\[m=\frac{(m/s)^2}{m/s^2}=\frac{\frac{m^2}{s^2}}{\frac{m}{s^2}}\]\[m=\frac{m^2}{\cancel{s^2}}*\frac{\cancel{s^2}}{m}=m\]

Answer 4

thanks a lot