Question

. A ball is dropped from the top of a tower 80m high at the same instant a second ball is thrown vertically upward from the ground. If the two balls meet each other at the point of 44m above the ground, determine the initial velocity of the second ball. 

Answer 1

Times required to meet the first ball with the second ball

$h=\frac{1}{2}gt^2$

$(80-44)=\frac{1}{2}\times 9.81\times t^2$

$t =2.7 sec$

the initial velocity of the second ball

$S = ut-\frac{1}{2}gt^2$

$44 = u\times 2.7-\frac{1}{2}\times 9.81\times 2.7^2$

$u = 29.54 \frac {m}{sec}$