Question

A BODY FALLS FROM A HEIGHT OF 45m ABOVE THE GROUND. FIND THE TIME TAKEN BY THE BODY TO REACH THE GROUND

Answer 1

the intial velocity is zero
as it is free fall u=0
distance s=1/2*g*t^2
45(2)=9.8t^2
$t^{2}= \frac{45(2)}{9.8}= \frac{90}{9.8}=9.1834 \\ t=3.03s$

Answer 2

Using the kinematic equation: s = ut + at²/2

Here, s = 45 m ;
u = 0 m/s 
a = 9.8 ms⁻²
t(in relation with u) = 0 s

Equating,
45 m = (0m/s × 0s) + (9.8 m/s²×t²)/2
⇒45 m = 9.8t²/2
⇒90 = 9.8t²
⇒9.183=t²
⇒t=3.030457 s