Question

A stone is dropped from a height of 19.6 m calculate the time taken to fall how fast does it move at the end of this fall acceleration after 1 second

Answer 1

Hope this helps you

Answer 2

now where the accelration is constant we can use formula

s=ut+1/2at^2

here

s= distance

u= initial velocity

t=time

19.6 = 9.8/2 * t^2

t = 2 sec

next part;

v = u +at

v= final velocity

v= 0+ 9.8* 1

= 9.8 m/s velocity after 1 sec.