A stone is dropped off a cliff, the last 1.5 seconds before it reaches the ground it travels 30 meters. Find the height from which it dropped & final velocity
distance during that
time=v'*1.5 -4.9*1.5^2
-30=v'*1.5-11.025
v'=-20+7.25=-14.75 m/s
WEll, you know that in each second and a half, the velocity increases by 9.8*1.5 m/s
so v' must represent the 14.75/(9.8*1.5)=.5 second after dropping, so it was dropped 2 seconds before it hit the ground, and it fell...
d=1/2 g t^2
check this, I did most of it in my head.
V'= -12.75m/s not -14.75 m/s
14.75/(9.8*1.5)≠.5
To find the height from which the stone was dropped and its final velocity, we can use the equations of motion under constant acceleration.
Let's define the variables:
- h: initial height (what we want to find)
- g: acceleration due to gravity (approximately 9.8 m/s²)
- t: time for the last 1.5 seconds of the fall (1.5 seconds)
- d: distance traveled during the last 1.5 seconds (30 meters)
- v: final velocity of the stone (what we want to find)
First, we can use the equation of motion for distance traveled:
d = v₀t + 0.5at²
Since the stone is dropped (initial velocity is zero), the equation simplifies to:
d = 0.5gt²
Substituting the known values:
30 = 0.5 * 9.8 * (1.5)²
Simplifying this equation, we find:
30 = 0.5 * 9.8 * 2.25
Next, we can solve for t:
67.5 = 4.425 * t²
Dividing both sides by 4.425:
t² = 15.306
Taking the square root of both sides:
t ≈ 3.91 seconds
Now that we know the total time of fall, we can find the initial height (h) of the stone using the equation:
h = 0.5gt²
Substituting the known values:
h = 0.5 * 9.8 * (3.91)²
Calculating this equation, we find:
h ≈ 73.7 meters
Therefore, the stone was dropped from a height of approximately 73.7 meters.
To find the final velocity (v), we can use the equation of motion:
v = v₀ + at
Since the stone was dropped (initial velocity is zero), the equation simplifies to:
v = gt
Substituting the known values:
v = 9.8 * (3.91)
Calculating this equation, we find:
v ≈ 38.26 m/s
Therefore, the final velocity of the stone is approximately 38.26 m/s.
To find the height from which the stone was dropped and its final velocity, we can use the equations of motion for free-falling objects.
The first equation relates the final velocity (v) of the stone to its initial velocity (u), acceleration due to gravity (g), and time taken (t):
v = u + gt
In this case, we are assuming that the stone was dropped from rest. Therefore, the initial velocity (u) is zero. The acceleration due to gravity (g) is a constant value of approximately 9.8 m/s^2.
So, the equation simplifies to:
v = gt
We know that the final velocity (v) is zero because the stone comes to a stop when it reaches the ground. Plugging in these values, we get:
0 = 9.8 * t
Solving for t, we find that t = 0 seconds.
Since the stone takes 1.5 seconds to travel the last 30 meters, we can set up another equation to find the time taken for the stone to reach the ground:
s = ut + (1/2) * gt^2
Here, s represents the distance traveled by the stone, u is the initial velocity (which is zero), and g is the acceleration due to gravity.
Plugging in the given values, we get:
30 = 0 + (1/2) * 9.8 * t^2
Multiplying through by 2 and dividing by 9.8, we find:
t^2 = 6.12
Taking the square root of both sides, we get:
t = 2.47 seconds
Therefore, it took the stone 2.47 seconds to fall from the height of the cliff.
Now, we can find the height (h) from which the stone was dropped using the equation:
h = ut + (1/2) * gt^2
Again, u is the initial velocity (which is zero), g is the acceleration due to gravity, and t is the time taken (2.47 seconds).
Plugging in the values, we get:
h = 0 + (1/2) * 9.8 * (2.47)^2
Simplifying this equation, we find:
h = 30.09 meters
Therefore, the stone was dropped from a height of approximately 30.09 meters.
To summarize:
- The stone was dropped from a height of approximately 30.09 meters.
- Its final velocity when it reaches the ground is zero.