Question

A stone is thrown horizontally from a tower. In $0.5$ second after the stone began to move, the numerical value of its velocity was $5 / 4$ times its initial velocity. The initial velocity of stone is $v \,m / s$. Determine $3\, v$

Answer 1

After $0.5\, \sec$ along vertical using $v _{ y }=u_{ y }+ a _{ y } t$
$v _{ y }=0+ g (0.5)=0.5\, g$
Along horizontal,
$v_{x}=v$
Using $v _{\text {net }} =\sqrt{ v _{ x }{ }^{2}+ v _{ y }{ }^{2}} \sqrt{ v ^{2}+0.25 g ^{2}}$
$=1.5\, v$
$v ^{2}+0.25 g ^{2}=2.25 v ^{2}$
$\Rightarrow 1.25 v ^{2}=(4.9)^{2}$
$v =4.382\, m / s \approx 4.4\, m / s$