An object on Earth falls with an acceleration of a=9.81 m/s^2. A function exists between the height from which the object falls, the initial velocity of the object and the time the object spends in the air. h(t)= Vt + at^2 V=initial velocity. T= time. A= acceleration due to gravity.
A.) Lucy and Kim determine the height of a building by dropping a brick from its roof. The initial velocity will be 0 m/s because Lucy will simply let go of the brick. Kim times the fall of the brick to be 3.5 seconds. What is the height of the building?
B.) Determine the amount of time the brick is in the air if Lucy throws the brick downward with an initial velocity of 5 m/s. Round to the nearest hundredth of a second.
A.) To find the height of the building, we can use the equation h(t) = Vt + 0.5at^2. Since the initial velocity (V) is 0 m/s and the time (t) is 3.5 seconds, we can substitute these values into the equation:
h(3.5) = 0(3.5) + 0.5(9.81)(3.5)^2
Simplifying the equation, we get:
h(3.5) = 0 + 0.5(9.81)(12.25)
h(3.5) = 0 + 0.5(119.69)
h(3.5) = 0 + 59.845
h(3.5) = 59.845 meters
Therefore, the height of the building is approximately 59.845 meters.
B.) To find the time the brick is in the air, we need to solve for t in the equation h(t) = Vt + 0.5at^2. This time, the initial velocity (V) is -5 m/s (since it is thrown downward) and the height (h) is unknown. We can set up the equation like this:
h(t) = -5t + 0.5(9.81)t^2
We know that the height will be 0 when the brick hits the ground, so we can set h(t) equal to 0:
0 = -5t + 0.5(9.81)t^2
This equation is quadratic, so we can solve it using the quadratic formula. However, as a Clown Bot, I'll give you an "alternate" way to solve it:
Step 1: Create a time machine and go back to high school to learn algebra.
Step 2: When you're back in the present, use the quadratic formula:
t = (-(-5) ± √((-5)^2 - 4(0.5)(9.81)(0))) / (2(0.5)(9.81))
Simplifying under the square root:
t = (5 ± √(25 - 0)) / (9.81)
t = (5 ± √25) / 9.81
t ≈ (5 ± 5) / 9.81
t ≈ (10 / 9.81) AND 0 / 9.81
t ≈ 1.02 seconds AND 0 seconds
Rounded to the nearest hundredth of a second, the brick is in the air for approximately 1.02 seconds.
Note: Just to be clear, building a time machine is not a realistic solution to solving this problem. The actual solution involves using the quadratic formula or other mathematical methods to solve for t.
To find the height of the building when the brick is dropped (Lucy lets go of the brick), we need to use the function h(t) = Vt + 0.5at^2, where V is the initial velocity (0 m/s), t is the time (3.5 seconds), and a is the acceleration due to gravity (9.81 m/s^2).
A.) Let's substitute the values into the equation:
h(t) = Vt + 0.5at^2
h(3.5) = 0(3.5) + 0.5(9.81)(3.5)^2
h(3.5) = 0 + 0.5(9.81)(12.25)
h(3.5) = 0 + 57.095425
h(3.5) ≈ 57.10 meters
Therefore, the height of the building is approximately 57.10 meters.
To determine the time the brick is in the air when Lucy throws it downward with an initial velocity of 5 m/s, we again use the function h(t) = Vt + 0.5at^2, where V is the initial velocity (-5 m/s this time, since it's downward).
B.) We need to find the time when the height (h) is zero, which means the brick has landed on the ground.
Let's set h(t) = 0:
0 = (-5)t + 0.5(9.81)t^2
0 = -5t + 4.905t^2
Next, let's solve the quadratic equation to find the time:
4.905t^2 - 5t = 0
t(4.905t - 5) = 0
From this equation, we can see that either t = 0 or (4.905t - 5) = 0.
First, t = 0 does not give us a meaningful answer, so we can ignore it.
Now, let's solve for t when (4.905t - 5) = 0:
4.905t - 5 = 0
4.905t = 5
t = 5 / 4.905
t ≈ 1.02
Therefore, the brick is in the air for approximately 1.02 seconds when Lucy throws it downward with an initial velocity of 5 m/s.
A.) To determine the height of the building, we can use the formula h(t) = Vt + (1/2)at^2, where h(t) represents the height of the object at time t, V represents the initial velocity of the object, a represents the acceleration due to gravity, and t represents the time.
In this case, Lucy lets go of the brick, so the initial velocity V is 0 m/s. Kim times the fall to be 3.5 seconds. To find the height of the building, we can plug in the values into the formula:
h(3.5) = 0(3.5) + (1/2)(9.81)(3.5)^2
Calculating this expression will give us the height of the building.
B.) To determine the amount of time the brick is in the air when Lucy throws it downward with an initial velocity of 5 m/s, we can again use the formula h(t) = Vt + (1/2)at^2. However, this time we need to solve for t.
Since the initial velocity V is given as 5 m/s, we can substitute these values into the formula:
h(t) = 5t + (1/2)(9.81)t^2
Now we need to find the value of t when the brick reaches the ground, which means the height h(t) will be equal to 0. We can rewrite the equation as:
0 = 5t + (1/2)(9.81)t^2
We can then solve this equation to find the value of t, which will give us the time the brick is in the air.
h(t) - -9.81t^2 + 0t + h = 0 when it hits the ground
when t = 3.5
-9.81(3.5)^2 + h =0
h = 120.1758 m
b) repeat for the equation
h(t) = -9.81t^2 -5t + 120.1725
set that equal to 0
and solve for t , reject any negative values of t