A large building has an inclined roof. The length of the roof is 64.0 m and the angle of the roof is 20.0° 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 3.35 m/s2. After leaving the edge of the roof, it falls a vertical distance of 33.5 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?

Why did the hammer go to school?

To get smarter! But in this case, it's going to get faster too! Let's calculate its journey down the inclined roof.

(a) To find the time it takes for the hammer to fall from the edge of the roof to the ground, we can first use the horizontal distance traveled by the hammer. The horizontal component of the 64.0 m inclined roof can be found using the formula:

horizontal distance = length of roof * cos(angle)
horizontal distance = 64.0 m * cos(20.0°)

Now let's find this value.

Oops, I dropped my calculator! Let me calculate it in my clown brain.

64.0 m * cos(20.0°) = approximately 60.595 m

So, the horizontal distance the hammer travels is approximately 60.595 m.

Now, we can find the time it takes for the hammer to fall from the edge of the roof to the ground using the vertical distance. We'll use the formula:

vertical distance = initial vertical velocity * time + (1/2) * acceleration * time^2

Since the hammer starts from rest, its initial vertical velocity is 0 m/s. And the vertical distance is 33.5 m. The acceleration is the acceleration due to gravity, approximately 9.81 m/s^2.

33.5 m = 0 * t + (1/2) * 9.81 m/s^2 * t^2

Now, let me calculate this.

Uh-oh, I slipped on a banana peel! I will collect myself and calculate it.

33.5 m = 4.905 m/s^2 * t^2

Wait a minute, we can simplify it further.

33.5 m = 4.905 m/s^2 * t^2
6.836 s^2 = t^2

Taking the square root of both sides:

t ≈ 2.615 s

So, it takes approximately 2.615 seconds for the hammer to fall from the edge of the roof to the ground! Ta-da!

(b) Now, let's calculate how far horizontally the hammer travels from the edge of the roof until it hits the ground. We already calculated the horizontal distance of the inclined roof as approximately 60.595 m. And we know the time it takes for the hammer to fall, which is 2.615 seconds.

To find the horizontal distance traveled by the hammer, we'll use the equation:

horizontal distance = initial horizontal velocity * time

Since there's no horizontal force acting on the hammer, its initial horizontal velocity is constant. So, the horizontal distance can be calculated as:

horizontal distance = horizontal velocity * time

Now, let me calculate this for you.

Before I calculate, I need to throw a pie in the air. Ready?

horizontal distance = 60.595 m * 2.615 s

And the answer is:

horizontal distance ≈ 158.447 m

The hammer travels approximately 158.447 meters horizontally from the edge of the roof until it hits the ground! Time for a grand finale!

To solve this problem, we can break it down into two parts: the motion of the hammer along the inclined roof, and the vertical free fall motion after leaving the edge of the roof.

Part a) How much time does it take the hammer to fall from the edge of the roof to the ground?

1. Find the time it takes for the hammer to slide down the inclined roof:
- Use the equation for motion along an inclined plane: s = ut + 0.5at^2
where s is the distance, u is the initial velocity (which is 0), a is the acceleration, and t is the time.
- The distance along the inclined roof is equal to its length, so s = 64 m.
- The acceleration along the roof is given as 3.35 m/s^2.
- Rearrange the equation to solve for time: t = √(2s/a)

Calculation: t = √(2(64 m)/(3.35 m/s^2))

2. Find the time it takes for the hammer to fall vertically:
- Use the equation for free fall: s = 0.5gt^2
where s is the vertical distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
- The distance that the hammer falls vertically is given as 33.5 m.
- Rearrange the equation to solve for time: t = √(2s/g)

Calculation: t = √(2(33.5 m)/(9.8 m/s^2))

3. Add the times from step 1 and step 2 to find the total time: total time = time along inclined roof + time of vertical fall.

Calculation: total time = t (step 1) + t (step 2)

Part b) How far horizontally does the hammer travel from the edge of the roof until it hits the ground?

1. Use the equation for motion in the horizontal direction:
- The time for the horizontal motion is the same as the total time calculated in part a.
- The horizontal distance is determined by the constant velocity in the horizontal direction, which can be calculated using the angle of the roof.
- The horizontal velocity, v_x, is given by v_x = v_initial * cos(θ), where θ is the angle of the roof (20.0°).
- The horizontal distance, d, can be calculated using d = v_x * total time.

Calculation: v_x = 3.35 m/s * cos(20.0°)
d = v_x * total time (from part a)

Note: Make sure to convert the angle from degrees to radians when using trigonometric functions. In this case, cos(20.0°) should be calculated using the angle in radians.

I will now perform the calculations.

To solve this problem, we can break it down into two separate parts: the motion of the hammer along the inclined roof and its subsequent free fall.

(a) To determine the time it takes for the hammer to fall from the edge of the roof to the ground, we need to find the time it takes for the hammer to slide down the inclined roof first. We can use the equations of motion for uniformly accelerated motion along an incline to do this.

First, let's determine the initial velocity of the hammer as it starts from rest. Since the hammer slides down the entire length of the roof, the displacement along the inclined roof is equal to its length: s = 64.0 m.

The angle of the roof is given as 20.0° below the horizontal, which means the component of gravity acting down the incline is given by: g*sin(20.0°).

Using the equation of motion for displacement along an incline: s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time, we can rearrange the equation to solve for time (t):

64.0 m = 0t + (1/2)(3.35 m/s²)t²

We can solve this quadratic equation for t by rearranging it into standard form:

(1/2)(3.35 m/s²)t² = 64.0 m

1.675t² = 64.0 m

t² = (64.0 m) / 1.675

t² = 38.209

t ≈ √38.209

t ≈ 6.18 s

Therefore, it takes approximately 6.18 seconds for the hammer to slide down the inclined roof.

Now, we need to determine the time it takes for the hammer to fall from the edge of the roof to the ground, which includes the free-fall time. We know that the vertical distance the hammer falls is 33.5 m.

Using the equation of motion for free fall in the vertical direction: h = 1/2gt², where h is the vertical distance, g is the acceleration due to gravity, and t is the time, we can rearrange the equation to solve for time (t):

33.5 m = 1/2(9.8 m/s²)t²

67.0 m = 9.8 m/s²t²

t² = 67.0 m / 9.8 m/s²

t² = 6.84

t ≈ √6.84

t ≈ 2.62 s

Therefore, it takes approximately 2.62 seconds for the hammer to fall from the edge of the roof to the ground.

(b) To determine how far horizontally the hammer travels from the edge of the roof until it hits the ground, we can use the horizontal velocity attained during its time sliding down the inclined roof.

The horizontal velocity will remain constant throughout the motion, as there is no horizontal acceleration acting on the hammer. We can find the horizontal velocity using the equation: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s since the hammer starts from rest), a is the acceleration, and t is the time.

Since the hammer slides down the entire length of the roof, the time of sliding down the roof will be equal to the time it takes to fall from the edge of the roof to the ground, which is approximately 2.62 s as calculated earlier.

Using the equation above, we can find the horizontal velocity:

v = 0 m/s + (3.35 m/s²)(2.62 s)

v ≈ 8.78 m/s

Finally, to find the horizontal distance travelled by the hammer, we multiply the time of free fall (2.62 s) by the horizontal velocity (8.78 m/s):

Horizontal distance = (2.62 s)(8.78 m/s)

Horizontal distance ≈ 23.0 m

Therefore, the hammer travels approximately 23.0 meters horizontally from the edge of the roof until it hits the ground.

How fast is it going when it leaves the roof?

v = a T
d = .5 a t^2
64 = .5 * 3.35 * t^2
so t = 6.18 seconds on roof
v = 3.35 * 6.18 = 20.7 m/s leaving roof

Vertical velocity leaving roof = -20.7 sin 20 deg = -7.08 m/s
horizontal velocity roof to ground = 20.7 cos 20 = 19.5 m/s

do vertical problem first
Vi = -20.7
a = -9.8 so a/2 = -4.9
Hi = 33.5
0 = 33.5 - 20.7 t - 4.9 t^2
so
t^2 + 4.22 t - 6.84 = 0

t = [ -4.22 +/- 6.72 ]/ 2
= 1.25 seconds

distance horizontal = 19.5 t = 19.5*1.25 = 24.4 metars