Question
A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s [down]. The cliff is 28.4m
above the ground. Determine the velocity of the rock just before it hits the ground.
above the ground. Determine the velocity of the rock just before it hits the ground.
Answers
GPT 3.5
To solve this problem, we can use the kinematic equation for final velocity:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (3.61 m/s downwards)
a = acceleration due to gravity (9.8 m/s^2 downwards)
s = displacement (28.4 m downwards)
Plugging in the values into the equation:
v^2 = (3.61)^2 + 2(-9.8)(-28.4)
v^2 = 13.0321 + 550.24
v^2 = 563.2721
Taking the square root of both sides:
v = √563.2721
v ≈ 23.74 m/s
The velocity of the rock just before it hits the ground is approximately 23.74 m/s [downwards].
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (3.61 m/s downwards)
a = acceleration due to gravity (9.8 m/s^2 downwards)
s = displacement (28.4 m downwards)
Plugging in the values into the equation:
v^2 = (3.61)^2 + 2(-9.8)(-28.4)
v^2 = 13.0321 + 550.24
v^2 = 563.2721
Taking the square root of both sides:
v = √563.2721
v ≈ 23.74 m/s
The velocity of the rock just before it hits the ground is approximately 23.74 m/s [downwards].