Question

A rock is thrown straight down from a height of 18 m. If it takes 0.8 s to hit the ground, what was the rock's initial speed? Air resistance can be neglected. A 30.3 m/s B 18.6 m/s C 26.4 m/s D 0.0 m/s E 14.7 m/s

Answer 1

B. 18.6 m/s

Step-by-step explanation:

d = Vot + (1/2)gt^2

Where:

d = initial height (18 m)

Vo = initial velocity (unknown)

t = time taken to hit the ground (0.8 s)

g = acceleration due to gravity (approximately 9.8 m/s^2)

18 m = Vo(0.8s) + (1/2)(9.8m/s^2)(0.8s)^2

18 m = Vo(0.8s) + 3.136 m

Vo(0.8s) = 18 m - 3.136 m

Vo(0.8s) = 14.864 m

Vo = 14.864 m / 0.8 s

Vo ≈ 18.6 m/s

Therefore, the rock's initial velocity was approximately 18.6 m/s in the downward direction.