From the top of a tower 100m high a ball is dropped and at the same instant another ball is projected vertically upwards from the ground so that it just reaches the top of the tower.At what height do the two balls pass one another?

Now both the ball will reach to the point in same time t.

$forthefirstball\phantom{\rule{0ex}{0ex}}100-x=ut+1/2a{t}^{2}\phantom{\rule{0ex}{0ex}}u=0,a=g\phantom{\rule{0ex}{0ex}}100-x=4.9{t}^{2}-----\left(1\right)$

$forthesecondball\phantom{\rule{0ex}{0ex}}finalspeed=0\phantom{\rule{0ex}{0ex}}initialspeedis{u}^{\text{'}}\phantom{\rule{0ex}{0ex}}distnacetravelled=100m\phantom{\rule{0ex}{0ex}}{v}^{2}-{u}^{\text{'}2}=2as\phantom{\rule{0ex}{0ex}}a=-g\phantom{\rule{0ex}{0ex}}{u}^{\text{'}}=44.27m/s\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}thereforeu\mathrm{sin}gsecondequationofmotion\phantom{\rule{0ex}{0ex}}x=44.27t-4.9{t}^{2}---\left(2\right)$

Solve the equation (1) and (2) and you will get

t = 2.26 s

x = 74.98 m

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