Question

a stone is dropped from a tower 180 m high. how long does it take to reach the ground? what is the velocity of the stone when it touches the ground? [take g =10m/s2

Answer 1

u=0 m/s
s=180 m
g=10 m/s^2
Using third eqaution of motion,
v^2-u^2=2gs
=)v^2-0=2×10×180
=)v=root 3600
Hence,v=60 m/s
Hope it helps u.

Answer 2

s = + 180 m

g = + 10 ms^-2

u = 0

t = ?

v = ?

Using Equation of Gravitation

${\implies s = ut + \dfrac{1}{2}gt^{2}}$

${\implies 180 = 0 + \dfrac{1}{2}\times t^{2}}$

t^2 = 36

t = 6 second

Now

v^2 = u^2 + 2gs

v^2 = 0 + 2 × 10 × 180

v^2 = 3600

v = 60 ms^-1