You are fixing the roof of your house when a hammer breaks loose and slides down. The roof makes an angle of 65o∘ with the horiz
ontal, and the hammer is moving at 9.5 m/s when it reaches the edge. Assume that the hammer is moving from the top of the roof to its right edge.
What is the horizontal component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the x-component of the velocity is to the right and negative value if the x-component of the velocity is to the left.
What is the vertical component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the direction of the y-component of the velocity is upward and negative value if the y-component of the velocity is downward.
What is the horizontal component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the x-component of the velocity is to the right and negative value if the x-component of the velocity is to the left.
What is the vertical component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the direction of the y-component of the velocity is upward and negative value if the y-component of the velocity is downward.
1 answer:
Answer:
Explanation:
Given:
initial speed of the hammer when leaving the edge of the roof along the inclination of the roof,
inclination of the roof form horizontal,
- Since the hammer is moving from the top of the roof to the right edge, its horizontal component will be towards right and vertical component will be towards downward direction.
Now the horizontal velocity:
<u>The vertical velocity:</u>
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1) See graph in attachment
2) 10 s
3) 50 m
Explanation:
1)
In this problem, we have an object initially moving with a velocity of
v = 10 m/s
when the time is
t = 0 s
Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.
The result can be found in the graph in attachment.
Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to
which is equivalent to the gradient of the line in the velocity-time graph.
2)
In this part, we want to find after what time the body will stop its motion.
To do that, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
In this problem:
u = 10 m/s is the initial velocity of the body
is the acceleration
v = 0 m/s, because we want to find the time T at which the body will stop
Re-arranging the equation, we find:
3)
In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
In this problem:
u = 10 m/s is the initial velocity
is the acceleration
t = 10 s is the time it takes for the body to stop (found in part 2)
Solving for s, we find the distance covered:
