A ball thrown up vertically returns to the thrower after 6s. Find:
(i) the velocity with which it was thrown up,
(ii) the maximum height it reaches, and
(iii) its position after 4s.
Numerical based on gravitation
Time of flight
t =2 X V/g
==> v=tg/2=6X10/2=25.85 m/sec
3) At the end of third second projectile researches it s amx height
Distance travelled with initial velocity o and acceleration g
is s=0.5 gt^2=0.5 mt
So projectile is 33.4-5=28.4 m from the ground.