Question

A stone is dropped from a height of 45m. Calculate it's velocity just before it touches the ground. Take g-10m/s²

Answer 1

since it is dropped then initial velocity u=o s=45 g=10

hence

v*v=u*u +2as

v*v= 2×10×45

v=30

Answer 2

s=45m

u=0

g=-10m/s 2

v=?

${v }^{2} = {u}^{2} + 2as \\ {v}^{2} = 0 + 2 \times 10 \times 45 \\ {v}^{2} = 900 \\ v = \sqrt{ 900}$

v = 30m/s