A rock is dropped off a 20m tall cliff. How long does it take to get to the bottom of the cliff? What is the instantaneous velocity right before it hits the bottom of the cliff?
yes
a. h = 0.5g*t^2
h = 20 m
g = +9,8 m/s^2
Solve for t.
b. V^2 = Vo^2 + 2g*h
Vo = 0
g = +9.8 m/s^2
h = 20 m.
Solve for V.
2 seconds?
To calculate the time it takes for the rock to reach the bottom of the cliff, we can use the kinematic equation for falling objects. The equation is:
d = (1/2) * g * t^2
Where:
- d is the distance (20m in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
- t is the time it takes to reach the bottom
We need to solve for t. Rearranging the equation, we have:
t^2 = (2 * d) / g
Taking the square root of both sides, we obtain:
t = √(2 * d / g)
Now let's substitute the given values:
t = √(2 * 20m / 9.8 m/s^2)
Calculating this, we find:
t ≈ √(40 / 9.8) ≈ 2.02 seconds (rounded to two decimal places)
So, it takes approximately 2.02 seconds for the rock to reach the bottom of the cliff.
To determine the instantaneous velocity right before the rock hits the bottom of the cliff, we can use the equation:
v = g * t
where:
- v is the velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
- t is the time (2.02 seconds in this case)
Substituting the values, we have:
v = 9.8 m/s^2 * 2.02 s
Calculating this, we find:
v ≈ 19.8 m/s (rounded to two decimal places)
Therefore, the instantaneous velocity right before the rock hits the bottom of the cliff is approximately 19.8 m/s.