Question

A rock is thrown upward from the hand at an initial speed of 4.5 m/s. How much later must a second rock be dropped from the same initial height of 50 cm so that the two rocks hit the ground at the same time?

Answer 1

for 1st rock:

vi = 4.5 m/s

a = -9.8 m/s^2

d = -50 cm

= -0.50 m

use:

d = vi*t + 0.5*a*t^2

-0.50 = 4.5*t + 0.5*(-9.8)*t^2

4.9*t^2 - 4.5*t - 0.50 = 0

This is quadratic equation (at^2+bt+c=0)

a = 4.9

b = -4.5

c = -0.5

Roots can be found by

t = {-b + sqrt(b^2-4*a*c)}/2a

t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 30.05

roots are :

t = 1.02 and t = -0.1002

since t can't be negative, the possible value of t is

t = 1.02 s

for 2nd rock:

vi = 0 m/s

d = -0.50 m

a = -9.8 m/s^2

use:

d = vi*t + 0.5*a*t^2

-0.50 = 0 - 0.5*9.8*t^2

t = 0.32 s

time difference = 1.02 s = 0.32 s

= 0.70 s

Answer: 0.70 s