A large building has an inclined roof. The length of the roof is 69.5 m and the angle of the roof is 18.0° below horizontal. A worker on the roof lets go of a hammer from the peak of the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 3.03 m/s2. After leaving the edge of the roof, it falls a vertical distance of 20.0 m before hitting the ground.
(a) How much time does it take the hammer to fall from the edge of the roof to the ground?
(b) How far horizontally does the hammer travel from the edge of the roof until it hits the ground?
To solve this problem, we need to calculate the time it takes for the hammer to fall from the edge of the roof to the ground, as well as the horizontal distance it travels during that time.
Let's start by finding the time it takes for the hammer to fall from the edge of the roof to the ground.
Step 1: Calculate the vertical distance traveled by the hammer.
Since the hammer falls a vertical distance of 20.0 m, this will be the distance we use for calculations.
Step 2: Use the kinematic equation to find the time of fall.
In this case, we'll use the equation for vertical motion:
Δy = v0t + (1/2)at^2
Where:
Δy = vertical distance (20.0 m)
v0 = initial velocity (0 m/s since it starts from rest)
a = acceleration (acceleration due to gravity = -9.8 m/s^2, negative because it opposes the motion)
t = time of fall (what we're trying to find)
Rearranging the equation to solve for t, we get:
t^2 + (2a)t - (2Δy/v0) = 0
Substituting the known values, we have:
t^2 + (2(-9.8))t - (2(20.0)/0) = 0
Simplifying further, we have:
t^2 - 19.6t - 0 = 0
Since this is a quadratic equation, we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -19.6, and c = 0. Plugging in these values, we get:
t = (-(-19.6) ± √((-19.6)^2 - 4(1)(0))) / (2(1))
Simplifying further, we have:
t = (19.6 ± √(384.16)) / 2
t ≈ (19.6 ± 19.6) / 2
Taking the positive root, we get:
t ≈ 19.6 / 2
t ≈ 9.8 seconds
Therefore, it takes approximately 9.8 seconds for the hammer to fall from the edge of the roof to the ground.
Now let's move on to calculating the horizontal distance traveled by the hammer.
Step 3: Calculate the horizontal distance traveled.
To find the horizontal distance, we need to find the horizontal velocity of the hammer when it reaches the edge of the roof.
Since the hammer slides down the entire length of the roof, the vertical component of its motion is zero when it reaches the edge. This means the hammer's final velocity (Vf) in the vertical direction is zero.
We can use the equation Vf = Vo + at in the vertical direction to find the time it takes for the hammer to reach the edge of the roof.
Vo = 0 (initial vertical velocity)
a = 3.03 m/s^2 (constant acceleration)
Using Vf = Vo + at, we have:
0 = 0 + (3.03)t
Solving for t, we get:
t = 0 s
This means it takes 0 seconds for the hammer to reach the edge of the roof.
Step 4: Calculate the horizontal distance.
To find the horizontal distance traveled by the hammer, we use the equation:
d = Vx * t
Where:
d = horizontal distance (what we're trying to find)
Vx = horizontal velocity (what we need to calculate)
t = time of fall (9.8 seconds)
Since there are no external horizontal forces acting on the hammer once it leaves the roof, its horizontal velocity remains constant. Therefore, the horizontal velocity at the edge of the roof (Vx) is the same as the horizontal velocity at the point of release (Vx0).
Using the equation Vf = Vo + at, but in the horizontal direction this time, we have:
Vx = Vx0 + axt
Since the hammer starts from rest, Vx0 = 0 and ax = 0.
Therefore, Vx = 0 + 0 * t
Vx = 0
This means the horizontal velocity of the hammer is 0 m/s.
Using the equation d = Vx * t, we have:
d = 0 * 9.8
d = 0
Therefore, the hammer does not travel any horizontal distance after leaving the edge of the roof until it hits the ground.
To summarize:
(a) The time it takes the hammer to fall from the edge of the roof to the ground is approximately 9.8 seconds.
(b) The hammer does not travel any horizontal distance from the edge of the roof until it hits the ground.