Question

A ball is thrown straight up from 3m above the ground with a velocity of 14m/s when does it hit the ground

Answer 1

Answer:

u=14

v=0(when the ball reaches the top point)

a= -10(gravity)

v=u+at

v-u/a=t

hence t=1.4 seconds

now distance travelled till top=v^2-u^2=2as

hence on solving s=9.8 metres

now when the ball comes down

u=0

a=10

s=10

hence s=ut + 1/2 at^2

or you can use the 3rd eq. of motion but the 3rd eq. will take the process to be a bit lenghty as after

applying 3rd eq. you will need to apply the 1st eq.

hence after solving (1) we have

t=1.4 seconds

therefore total time=1.4+1.4=2.8 seconds