Question

A ball is thrown vertically upward with a speed of 10 M per second from the top of a tower 200 metres high and another is thrown vertically downwards with the same speed simultaneously. the time difference between them in reaching the ground in seconds(g=10 m/second square) is? this question is taken from the chapter of kinematics, topic vertical motion.

Answer 1

when ball a is thrown upwards.
y1=ut-1/2gt^2
y1=10t-5t^2

while for ball b.
y2=-10t-5t^2

when y1=-200
-200=10t-5t^2
t^2-2t-40=0
t=(2+rootof(4+160))/2

when y2=-200
then quadratic eqn becomes.
t^2+2t-40=0
then
t=(-2+rootof(4+160))/2

so diff. of time taken by both balls =
(2+root......)/2-(-2+root....)/2
=1+1=2sec

*** i have solved quadratic eqn. i neglected one answer in each eqn because that're imaginary.
if you kno how to solve quad. eqn. otherwise you can ask me.