a stone is projected upward with a speed of 80.0ft/s from a 224 ft tower. find the maximum height and the speed at which it strikes the ground.
Assuming no air resistance, and PE at bottom of tower to be zero, then
At the maximum height,
mgh=(224mg+m(80²)/2
h=224+80²/(2g) [assume g=32.2 '/s²
=224'+99.38'
=323.38'
At bottom of tower,
mv²/2 = m(80²)/2+mg(224)
solve for v to get
v=144.31 '/s
Why did the stone want to reach the maximum height? Because it wanted to be the "peak" performer! Now, let's calculate the maximum height and the speed at which it strikes the ground.
To find the maximum height, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (0 ft/s at the maximum height)
u = initial velocity (80.0 ft/s upwards)
a = acceleration due to gravity (-32.2 ft/s², considering upward as positive)
s = displacement (maximum height)
Plugging in the values, we get:
0² = (80.0)² + 2(-32.2)s
Solving for s, we get:
s = (80.0)² / (2 * 32.2) = 101.24 ft
Therefore, the maximum height reached by the stone is approximately 101.24 ft.
Now, to find the speed at which it strikes the ground, we use the equation:
v = u + at
Where:
v = final velocity (speed at which it strikes the ground)
u = initial velocity (0 ft/s at the maximum height)
a = acceleration due to gravity (-32.2 ft/s², considering downward as negative)
t = time taken to reach the ground
We can find the time taken to reach the ground using the equation:
s = ut + (1/2)at²
Where:
s = displacement (224 ft downward)
u = initial velocity (0 ft/s at the maximum height)
a = acceleration due to gravity (-32.2 ft/s², considering downward as negative)
t = time taken to reach the ground
Plugging in the values, we get:
224 = 0 * t + (1/2)(-32.2)t²
Simplifying, we have:
-16.1t² = 224
Dividing by -16.1, we get:
t² = -224 / -16.1 = 13.91
Taking the square root, we find:
t ≈ 3.73 s
Now, plugging this time value into the equation v = u + at, we get:
v = 0 + (-32.2)(3.73) = -120.07 ft/s
Note: The negative sign indicates the downward direction.
Therefore, the speed at which the stone strikes the ground is approximately 120.07 ft/s (when rounded to two decimal places).
Remember, though, that all these calculations conveniently ignore air resistance because it can be a real "drag"!
To find the maximum height the stone reaches, we can use the kinematics equation for vertical motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 ft/s at the highest point)
u = initial velocity (80.0 ft/s)
a = acceleration due to gravity (-32.2 ft/s^2, assuming upward direction as positive)
s = displacement (maximum height)
Substituting the given values into the equation, we can find the maximum height (s):
0 = (80.0 ft/s)^2 + 2(-32.2 ft/s^2) * s
Rearranging the equation:
(80.0 ft/s)^2 = 2 * 32.2 ft/s^2 * s
Simplifying the equation further:
s = (80.0 ft/s)^2 / (2 * 32.2 ft/s^2)
Now we can calculate the maximum height:
s = 12800 ft^2/s^2 / 64.4 ft/s^2
s ≈ 198.14 ft
Therefore, the maximum height reached by the stone is approximately 198.14 ft.
To find the speed at which it strikes the ground, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (0 ft/s when falling from the maximum height)
a = acceleration due to gravity (-32.2 ft/s^2, assuming downward direction as positive)
t = time taken to reach the ground
Substituting the given values into the equation:
v = 0 ft/s + (-32.2 ft/s^2) * t
To find t, we can use the kinematic equation:
s = ut + (1/2)at^2
Where:
s = initial displacement (198.14 ft)
u = initial velocity (0 ft/s)
a = acceleration due to gravity (-32.2 ft/s^2)
Substituting the given values into the equation:
198.14 ft = 0 ft/s * t + (1/2)(-32.2 ft/s^2) * t^2
Simplifying the equation:
-16.1 ft/s^2 * t^2 = 198.14 ft
Solving for t by isolating t^2:
t^2 = 198.14 ft / (-16.1 ft/s^2)
t ≈ 4.89 s (taking the positive value since time cannot be negative)
Now we can calculate the final velocity (v) by using the equation:
v = 0 ft/s + (-32.2 ft/s^2) * 4.89 s
v ≈ -157.56 ft/s
Therefore, the speed at which the stone strikes the ground is approximately 157.56 ft/s. Note the negative sign indicates a downward velocity.
To find the maximum height reached by the stone and the speed at which it strikes the ground, we can use the equations of motion for linear motion under gravity.
First, let's determine the maximum height reached by the stone. We can use the equation:
h = (v^2 - u^2) / (2g)
where:
h = maximum height
v = final velocity (0 ft/s at maximum height)
u = initial velocity (80.0 ft/s)
g = acceleration due to gravity (32.2 ft/s^2)
Plugging in the values, we get:
h = (0^2 - 80.0^2) / (2 * 32.2)
h = (-6400.0) / 64.4
h ≈ -99.38 ft
Since height cannot be negative in this context, the maximum height reached by the stone is approximately 99.38 ft.
Next, let's calculate the time it takes for the stone to hit the ground. We can use the equation:
s = ut + (1/2)gt^2
where:
s = distance traveled (224 ft)
u = initial velocity (80.0 ft/s)
g = acceleration due to gravity (32.2 ft/s^2)
t = time taken
Rearranging the equation to solve for t:
32.2t^2 + 80.0t - 224 = 0
Now we can solve this quadratic equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where:
a = 32.2
b = 80.0
c = -224
After substituting the values into the quadratic formula and simplifying, we find two possible values for t:
t ≈ 1.98 s or t ≈ -3.56 s
We discard the negative value since time cannot be negative. Therefore, it takes approximately 1.98 seconds for the stone to hit the ground.
Finally, let's find the speed at which the stone strikes the ground. We can use the equation:
v = u + gt
where:
v = final velocity
u = initial velocity (80.0 ft/s)
g = acceleration due to gravity (32.2 ft/s^2)
t = time taken (1.98 s)
Plugging in the values, we get:
v = 80.0 + (32.2 * 1.98)
v = 80.0 + 63.636
v ≈ 143.636 ft/s
Thus, the speed at which the stone strikes the ground is approximately 143.636 ft/s.