A person stands on top of a tall building holding two rocks at a height of 50 meters. At the same moment, one rock is dropped from rest while the other is thrown downwards with an initial velocity of 8.0 m/s. Neglecting air resistance, calculate how long after the first rock lands does the second rock strike the ground

Answer

The second rock will land 2.4s after the first rock

Explanation:

Given that

Height of the building s=50m

We assume that the first rock is acting with gravity so that a=9.81m/s

And initial velocity u=0

Applying the equation of motion

S=ut+1/2at²

50=0*t+1/2(9.81)t²

50=4.905t²

t²=50/4.905

t²=10.19

t=√10.19

t=3.19sec

For the second rock initial velocity u=8m/s and v=0 and a=9.81

Applying the equation of motion

v=u+at

0=8+9.81t

t=-8/9.81

t=0.81sec

Hence the second rock will land 2.4s after the first rock

I.e

3.19-0.81

answer: pseudoscience