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A ball is thrown horizontally from the top of a tower with a velocity of 40 ms-1. Take g = 10 ms-2

a) Find the horizontal and vertical displacement after 1,2,3,4,5 seconds, then the path of the motion of ball.

b) If the ball reaches the ground in 4 seconds, find the height of the tower.

For horizontal distance, gravity plays no role.

$HorizontalDis\mathrm{tan}ce=vt\phantom{\rule{0ex}{0ex}}So,in1,2,3,4,5secthedis\mathrm{tan}cewillbe\phantom{\rule{0ex}{0ex}}40m,80m,120m,160m,200m\phantom{\rule{0ex}{0ex}}Forverticaldis\mathrm{tan}ce\phantom{\rule{0ex}{0ex}}S=ut+0.5g{t}^{2}\phantom{\rule{0ex}{0ex}}For1sec,s=45m\phantom{\rule{0ex}{0ex}}for2sec\phantom{\rule{0ex}{0ex}}S=80+20=100m\phantom{\rule{0ex}{0ex}}For3sec\phantom{\rule{0ex}{0ex}}S=120+45=165m\phantom{\rule{0ex}{0ex}}For4sec\phantom{\rule{0ex}{0ex}}S=160+80=240m\phantom{\rule{0ex}{0ex}}For5sec\phantom{\rule{0ex}{0ex}}S=200+125=325m\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The path of the ball will be parabolic

(b) When the ball hits the ground, its final velocity is zero

$H=\frac{{u}^{2}}{2g}\phantom{\rule{0ex}{0ex}}H=\frac{{40}^{2}}{20}=80m$

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