It takes 3 s for a rock to hit the ground when it is thrown straight up from a cliff with an initial velocity of 8.63 m/s. how long a time would it take to reach the ground if it is thrown straight down with the same speed?

Explanation:

h = height of the cliff

Consider upward direction as positive and downward direction as negative

Consider the motion of rock thrown straight up :

Y = vertical displacement = - h

v₀ = initial velocity = 8.63 m/s

a = acceleration = - 9.8 m/s²

t = time taken to hit the ground = 3 s

Using the equation

Y = v₀ t + (0.5) a t²

- h = (8.63) (3) + (0.5) (- 9.8) (3)²

h = 18.21 m

Consider the motion of rock thrown down :

Y' = vertical displacement = - 18.21

v'₀ = initial velocity = - 8.63 m/s

a' = acceleration = - 9.8 m/s²

t' = time taken to hit the ground = ?

Using the equation

Y' = v'₀ t' + (0.5) a' t'²

- 18.21 = (- 8.63) t' + (0.5) (- 9.8) t'²

t' = 1.2 s

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge