Question

A ball at rest heart rolling down a hill with a constant celebration of 3.2 m/s squared what is the final velocity of the ball after 6.0 seconds

Answer 1

Answer:

19.2m/s

Explanation:

Given=

initial velocity(u)=0, time(t)=6 sec, acceleration=3.2m/s^2

now using first equation of motion

v=u+at

v=0+3.2*6

v=0+19.2

v=19.2m/s

Answer 2

Answer:

a=v-u/t

then,3.2=v-0/6

by taking 6 on L.H.S

3.2*6=v

19.2=final velocity

thanks