If it takes 2.37 seconds for a stone to hit the ground from a cliff. (a) From what height was the stone dropped from? (b) What is the stones final velocity when it hits the ground?
vfinal=g*time
height=1/2 g t^2
To answer these questions, we can use the equations of motion under constant acceleration due to gravity. Let's break this down step by step.
(a) To determine the height from which the stone was dropped, we can use the equation:
h = (1/2) * g * t^2
where:
h is the height (unknown),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time taken to reach the ground (2.37 seconds).
Now let's substitute the values into the equation:
h = (1/2) * 9.8 * (2.37)^2
Calculating this, we find:
h ≈ (1/2) * 9.8 * 5.6169
≈ 27.7056 meters
Therefore, the stone was dropped from a height of approximately 27.7056 meters.
(b) To calculate the stone's final velocity when it hits the ground, we can use the equation:
v = g * t
where:
v is the final velocity (unknown),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time taken to reach the ground (2.37 seconds).
Plugging in the values:
v = 9.8 * 2.37
Calculating this, we find:
v ≈ 23.226 m/s
Therefore, the stone's final velocity when it hits the ground is approximately 23.226 m/s.