Jul 25, 2016

Given

$u \to \text{The initial upward velocity of the ball} = + 19.8 \frac{m}{s}$

$h \to \text{Downward displacement in t=5 s"= "height of tower}$

$g \to \text{Acceleration due to gravity} = - 9.8 \frac{m}{s} ^ 2$

Applying equation of motion

$h = u \cdot t + \frac{1}{2} g \cdot {t}^{2}$

$\implies h = 19.6 \cdot 5 - \frac{1}{2} \cdot 9.8 \cdot {5}^{2}$

$\implies h = 98 - 4.9 \cdot 25 = - 24.5 m$

Velocity of the ball on reaching the ground

$v = u + g \cdot t$

$\implies v = 19.6 - 9.8 \cdot 5 = 19.6 - 49$
$= - 29.4 \frac{m}{s}$

Negative sign indicates that direction of final displacement and velocity are opposite to the direction of initial velcity.