1. A ball is projected horizontally from the edge of a table that is 4.0 meters high, and strikes the floor at a point
3.0 meters from a point on the floor directly below the edge of the table. The initial velocity of the ball as it is projected from the table is, in m/s ( g=9.8 m/s2)
2. At an airport, luggage is loaded into two cars of a luggage carrier being pulled by a luggage motocart whose mass
is 1100 kg. The car connected to the motorcart has a mass of 300 kg and the last car has a mass of 400 kg. If the
luggage carrier is accelerating at 1.3 m/s2 , the the tension in the coupling bar that connects the first car to the
motorcart is, in Newtons,
3. A traffic light weighing 140 N hangs from a vertical cable which is fastened to two other upper cables which are
attached to a horizontal support. One of the upper cables makes an angle of 30 degrees with the horizontal support
and the other upper cable makes an angle of 55 degrees with the horizontal support. The tension in the upper cable
that makes an angle of 55 degrees with the horizontal support is, in Newtons
4. A soccer player kicks a soccer ball from a point 20.0 m from the goal line. The soccer ball leaves the ground
with a speed of 20.0 m/s at an angle of 35 degrees to the horizontal. The height of the soccer ball as it passes over
the goal line is, in meters (Assume g = 9.8 m/s2)
1. One question per post, please, and show your work. In the case of these four, you did not even provide the ___ where the answer is supposed to go.
In the case of #1, use the height to compute the fall time T. Then use
Vo * T = 3.0 meters,
where Vo is the initial velocity (which is horizontal) and 3.0 m is the horizontal distance travelled.
Thanks for the timely help
1. To find the initial velocity of the ball, we can use the equations of motion. In this case, the ball is projected horizontally, which means its initial vertical velocity is zero.
We can use the equation of motion for vertical motion to find the time it takes for the ball to reach the ground:
h = ut + (1/2)gt^2
Where:
h = height of the table = 4.0 m
u = initial vertical velocity = 0 m/s
g = acceleration due to gravity = 9.8 m/s^2
t = time
Substituting the values into the equation, we get:
4.0 = 0 + (1/2)(9.8)t^2
t^2 = (4.0 * 2) / 9.8
t^2 = 0.8163
t ≈ √0.8163
t ≈ 0.9047 s
Now, we can use this time to find the horizontal distance the ball travels. Since the ball is projected horizontally, its horizontal velocity remains constant throughout its motion.
We can use the equation of motion for horizontal motion:
s = ut
Where:
s = horizontal distance traveled = 3.0 m
u = initial horizontal velocity (which is the same as the final horizontal velocity) = v
Substituting the values into the equation, we get:
3.0 = v * 0.9047
v = 3.0 / 0.9047
v ≈ 3.31 m/s
Therefore, the initial velocity of the ball as it is projected from the table is approximately 3.31 m/s.
2. To find the tension in the coupling bar, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
F_net = m * a
Where:
F_net = net force
m = total mass of the luggage carrier (including the motocart and both cars)
a = acceleration of the luggage carrier
First, we need to find the total mass of the luggage carrier:
total mass = mass of motocart + mass of first car + mass of last car
total mass = 1100 kg + 300 kg + 400 kg
total mass = 1800 kg
Substituting the values into the equation, we get:
F_net = 1800 kg * 1.3 m/s^2
F_net ≈ 2340 N
Therefore, the tension in the coupling bar that connects the first car to the motocart is approximately 2340 N.
3. To find the tension in the upper cable that makes an angle of 55 degrees with the horizontal support, we can use the concept of vector resolution.
We can resolve the weight of the traffic light into its horizontal and vertical components. The vertical component will be balanced by the tension in the two upper cables, while the horizontal component will be zero.
Let's consider the cable that makes an angle of 55 degrees with the horizontal support. We can find its vertical component using trigonometry:
T * cos(55) = weight
T * cos(55) = 140 N
T = 140 N / cos(55)
Using a calculator, we can find that cos(55) ≈ 0.5736. Substituting the values into the equation, we get:
T ≈ 140 N / 0.5736
T ≈ 243.56 N
Therefore, the tension in the upper cable that makes an angle of 55 degrees with the horizontal support is approximately 243.56 N.
4. To find the height of the soccer ball as it passes over the goal line, we can use the projectile motion equations. In this case, we are given the initial velocity of the soccer ball and the angle of projection.
First, we can find the time it takes for the soccer ball to reach the goal line. We can use the equation of motion for horizontal motion:
s = ut
Where:
s = horizontal distance traveled = 20.0 m
u = initial horizontal velocity = v * cos(35)
Substituting the values into the equation, we get:
20.0 = (20.0 * cos(35)) * t
t = 20.0 / (20.0 * cos(35))
Using a calculator, we can find that cos(35) ≈ 0.8192. Substituting the values into the equation, we get:
t ≈ 20.0 / (20.0 * 0.8192)
t ≈ 20.0 / 16.384
t ≈ 1.2207 s
Now, we can use this time to find the height of the soccer ball at that time. We can use the equation of motion for vertical motion:
h = ut + (1/2)gt^2
Where:
h = height of the soccer ball
u = initial vertical velocity = v * sin(35)
g = acceleration due to gravity = 9.8 m/s^2
t = time
Substituting the values into the equation, we get:
h = (20.0 * sin(35)) * 1.2207 + (1/2)(9.8)(1.2207)^2
Using a calculator, we can find that sin(35) ≈ 0.5736. Substituting the values into the equation, we get:
h ≈ (20.0 * 0.5736) * 1.2207 + (1/2)(9.8)(1.2207)^2
Simplifying the equation, we get:
h ≈ 11.47 + 7.534
h ≈ 18.004
Therefore, the height of the soccer ball as it passes over the goal line is approximately 18.004 meters.