While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.05 m/s. The stone subsequently falls to the ground, which is 12.7 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.81 m/s2. (This is NOT a suggestion to carry out such an experiment!)
a. V^2 = Vo^2 + 2g*h1.
0 = 6.05^2 - 19.6h1, h1 = 1.87 m.
V^2 = Vo^2 + 2g*(h1+h2) = 0 + 19.6*(1.87+12.7) = 285.6, V = 16.9 m/s.
b. V = Vo + g*Tr.
0 = 6.05 - 9.8Tr, Tr = 0.617 s. = Rise time.
h = 0.5g*Tf^2.
1.87+12.7 = 4.9Tf^2, Tf = 1.72 s.
= Fall time.
Tr+Tf = 0.617 + 1.72 = 2.34 s. = Time in air.
To find the speed at which the stone impacts the ground, we can use the concept of conservation of energy. Initially, the stone is at a height of 0 m and has potential energy. When it reaches the ground, its potential energy becomes zero, and all the initial potential energy is converted into kinetic energy.
The potential energy at the topmost point is given by the formula:
PE = mgh, where m is the mass of the stone, g is the acceleration due to gravity, and h is the height.
Given that g = 9.81 m/s^2, h = 12.7 m, and m is not provided, we can ignore the mass since it cancels out when calculating the speed at which the stone impacts the ground.
PE = mgh = 0.5mv^2, where v is the velocity at which the stone impacts the ground.
Solving for v, we get:
v = √(2gh)
Substituting the values, we have:
v = √(2 * 9.81 * 12.7) ≈ 15.8 m/s
Therefore, the speed at which the stone impacts the ground is approximately 15.8 m/s.
To find the time the stone is in the air, we can use the kinematic equation:
h = ut + 0.5gt^2.
Here, h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
Since the stone is thrown upward and then falls downward, the total time in the air is twice the time it takes to reach the maximum height.
To find the time it takes to reach the maximum height, we use the equation:
u = v - gt,
where u is the initial velocity, v is the final velocity (0 m/s at the maximum height), g is the acceleration due to gravity, and t is the time.
0 = 6.05 - 9.81t.
Solving for t:
t = 6.05 / 9.81 ≈ 0.62 s.
Therefore, the time taken to reach the maximum height is approximately 0.62 seconds.
The total time in the air (falling downward) is twice that value:
Total time = 2 * 0.62 ≈ 1.24 seconds.
Therefore, the stone is in the air for approximately 1.24 seconds.
To solve this problem, we can break it down into two separate parts: the stone's upward motion and its downward motion.
First, let's find the time it takes for the stone to reach its maximum height. We can use the equation:
v = u + gt
where:
v = final velocity (0 m/s since the stone reaches its highest point)
u = initial velocity (6.05 m/s)
g = acceleration due to gravity (-9.81 m/s^2)
t = time
Rearranging the equation, we get:
t = (v - u) / g
Substituting the values into the equation, we have:
t = (0 - 6.05) / -9.81
Simplifying, we find:
t = 0.617 seconds
So it takes 0.617 seconds for the stone to reach its maximum height.
Next, we can find the time it takes for the stone to fall from its maximum height to the ground. We can use the equation:
s = ut + (1/2)gt^2
where:
s = distance (12.7 m)
u = initial velocity (0 m/s since the stone is momentarily at rest)
g = acceleration due to gravity (-9.81 m/s^2)
t = time
Rearranging the equation, we get a quadratic equation:
(1/2)gt^2 = s
Substituting the values into the equation, we have:
(1/2)(-9.81)t^2 = 12.7
Simplifying, we find:
-4.905t^2 = 12.7
Now, solving for t, we have:
t^2 = -12.7 / -4.905
t^2 = 2.59
t ≈ √2.59
t ≈ 1.61 seconds
So it takes approximately 1.61 seconds for the stone to fall from its maximum height to the ground.
Therefore, the total time the stone is in the air is the sum of the time to reach the maximum height and the time to fall to the ground:
Total time = 0.617 + 1.61 ≈ 2.23 seconds
Now, to find the speed at which the stone impacts the ground, we can use the equation:
v = u + gt
where:
v = final velocity (the speed at impact)
u = initial velocity (0 m/s since the stone is momentarily at rest)
g = acceleration due to gravity (-9.81 m/s^2)
t = time
Rearranging the equation, we have:
v = gt
Substituting the values into the equation, we have:
v = -9.81 * 1.61
v ≈ -15.8 m/s
Since the stone is falling downward when it impacts the ground, we consider the negative sign to indicate its direction. Thus, the speed at which the stone impacts the ground is approximately 15.8 m/s.