Question

pls......answer the 12th question in he paper....

Answer 1

@AravindG

Answer 2

@henpen can u help

Answer 3

use v= u + at .....where v = final velocity which is zero......( at max height)
u= initial velocity with which it was thrown......after ball is thrown only gravitational force acts on it ....so a= -g....we take negative sign as direction of motion of ball ( upwards} is opposite to the direction of force..( downwards)..
so you will get value if velocty......by putting t= 3 ....

when u get u , use s = ut - 1/2 g t^2 to find out max height reached.....
or u can also find max height using equation \[v ^{2}= u ^{2}- 2gh\]......by putting v=0