Question

A ball is dropped from rest at a height of 95m above the ground..... What is the speed before it touches the ground? How long does it take to reach the ground?

Answer 1

Data:-vi=0 ,h=95m , vf=? , t=?

Step-by-step explanation:

so here we have to find two things vf and then t so for finding vf applimg 3rd eq of motion 2gs=vf²-vi² , vf=2gs+vi² , vf= 2×9.8×95+(0)² , vf=1862 taking sq root on b.s

`\sqrt{vf {}^(2) } = √(1862)`

vf=43.14 now we have to find time so for t apply 1st eq of motion vf=vi+gt , t=vf-vi/g , t=43.14-0/9.8 , t= 4.403 (vf=43.14 and time=4.403)