Question #adb98

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Answer 1 · 2016-07-25T14:18:47

Given

$u \to The initial upward velocity of the ball = + 19.8 \frac{m}{s}$ $u->"The initial upward velocity of the ball"=+19.8m/s$

$h \to Downward displacement in t=5 s = height of tower$ $h->"Downward displacement in t=5 s"= "height of tower"$

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

Applying equation of motion

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

$\Rightarrow h = 19.6 \cdot 5 - \frac{1}{2} \cdot 9.8 \cdot 5^{2}$ $=>h=19.6*5-1/2*9.8*5^2$

$\Rightarrow h = 98 - 4.9 \cdot 25 = - 24.5 m$ $=>h=98-4.9*25=-24.5m$

Velocity of the ball on reaching the ground

$v = u + g \cdot t$ $v= u+g*t$

$\Rightarrow v = 19.6 - 9.8 \cdot 5 = 19.6 - 49$ $=>v=19.6-9.8*5=19.6-49$
$= - 29.4 \frac{m}{s}$ $=-29.4m/s$

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