Question

A ball is thrown vertically upwards with an initial velocity of $49m{s}^{-1}$. Calculate the time taken by it before it reaches the ground again. (Take $g=9.8m{s}^{-2}$)

Answer 1

Given,
Initial velocity, $u=49m{s}^{-1}$ and,
$a=-g=-9.8m{s}^{-2}$
As ball again return to ground, so $S=0$
Using second equation of motion for complete journey of ball,
$S=ut+\frac{1}{2}a{t}^2$
$0=49\times T+\frac{1}{2}\times (-9.8)\times T^2$
$49\times T=4.9\times T^2$
$T^2-10T=0$
$T(T-10)=0$
$\Rightarrow T=0.T=10$
$T=0$, is the initial time, hence total time for complete journey of ball is equal to $10s$.