1) A bartender slides a beer mug at 1.4 m/s towards a customer at the end of a frictionless bar that is 1.8 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar.
(a) How far away from the end of the bar does the mug hit the floor?
0.849m
Just confused on part b)
(b) What are the speed and direction of the mug at impact?
Speed
Direction (below the horizontal)
2) Suppose the ball is thrown from the same height as in the PRACTICE IT problem (45.0m) at an angle of 32.0° below the horizontal. If it strikes the ground 43.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
(b) the initial speed (m/s)
(c) the speed and angle of the velocity vector with respect to the horizontal at impact
speed (m/s)
angle (° below the horizontal)
1a. h = 0.5*g*t^2 s.= 1.8.
0.5*9.8*t^2 = 1.8,
t = 0.61 s. = Fall time.
d = V * t = 1.4 * 0.61 =
=
1b. Y^2 = Yo^2 + 2g*h = 0 + 19,6*1,8 = 35.3,
Y = 5.94 m/s.
V = sqrt(1,4^2 + 5.94^2) =
Tan A = Y/X = 5.94/1.4
A =
To solve part (b) of the first question, we can use the conservation of mechanical energy. Since there is no friction, the only form of energy involved is potential energy (mgh) and kinetic energy (1/2mv^2). We can equate the initial potential energy (when the mug leaves the end of the bar) with the final kinetic energy (when the mug hits the floor).
Initial potential energy = Final kinetic energy
mgh = 1/2mv^2
Since mass (m) is common on both sides of the equation, it cancels out:
gh = 1/2v^2
Rearranging the equation, we get:
v^2 = 2gh
Taking the square root of both sides, we get:
v = sqrt(2gh)
Plugging in the values given in the question, we have:
v = sqrt(2 * 9.8 m/s^2 * 1.8 m)
v = sqrt(35.28)
v ≈ 5.94 m/s
Therefore, the speed of the mug at impact is approximately 5.94 m/s.
For the direction, it is stated in the question that the mug is sliding off the end of the bar, which means it is moving horizontally. So the direction of the mug at impact is below the horizontal.
Now moving on to the second question:
To solve part (b), we need to use the equations of motion for projectile motion. Since the ball is thrown at an angle below the horizontal, we can split the initial velocity into horizontal and vertical components.
The horizontal component of the initial velocity (vx) remains constant, while the vertical component of the initial velocity (vy) changes due to acceleration from gravity.
Given that the ball strikes the ground 43.8 m away, we can use the equation for horizontal displacement to eliminate the product of initial velocity and time (v0t) from the equation for vertical displacement.
Horizontal displacement (x) = v0x * t
Vertical displacement (y) = v0y * t + (1/2) * g * t^2
Since the question asks for the initial speed (v0), we can use the equation for horizontal displacement:
x = v0 * cosθ * t
Rearranging the equation, we get:
v0 = x / (cosθ * t)
We are given the values for x (43.8 m) and θ (32° below the horizontal).
To calculate t, we can use the equation for vertical displacement:
y = v0 * sinθ * t - (1/2) * g * t^2
Since the ball strikes the ground, the vertical displacement will be equal to -45.0 m. Substituting the values, we have:
-45.0 m = v0 * sin(32°) * t - (1/2) * 9.8 m/s^2 * t^2
This is a quadratic equation in t. Solving for t, we get two possible values: t1 and t2.
To find the positive value of t, we can ignore the negative root (since time cannot be negative in this context).
Once we have the value of t, we can substitute it into the equation for v0:
v0 = x / (cosθ * t)
Solving for v0 will give us the initial speed of the ball.
Unfortunately, the part (a) of the second question is missing. Please provide the missing part (a) for a complete answer.
To find the speed and direction of the mug at impact in question 1b, we can use the principles of projectile motion.
The initial velocity of the mug is given as 1.4 m/s horizontally. Since there is no friction acting on the mug, its horizontal velocity remains constant throughout its motion.
To find the speed at impact, we need to find the vertical velocity of the mug just before it hits the ground. We can use the equation of motion:
vf^2 = vi^2 + 2ad
Where vf is the final velocity, vi is the initial velocity, a is the acceleration (in this case, due to gravity -9.8 m/s^2), and d is the vertical displacement (1.8 m - the height of the bar).
To find the vertical displacement, we can use the equation:
d = vit + (1/2)at^2
Where vi is the initial velocity in the vertical direction (0 m/s since it was not given), a is the acceleration (-9.8 m/s^2), t is the time it takes for the mug to hit the ground, and d is the vertical displacement.
We can rearrange the equation to solve for t:
t = sqrt(2d/a)
Substituting the values:
t = sqrt(2 * 1.8 m / 9.8 m/s^2)
t ≈ 0.428 s (rounded to three decimal places)
Now that we know the time it takes for the mug to hit the ground, we can find the vertical velocity just before impact using:
vf = vi + at
Where vi is the initial vertical velocity (0 m/s) and a is the acceleration (-9.8 m/s^2).
vf = -9.8 m/s * 0.428 s
vf ≈ -4.2 m/s (rounded to two decimal places)
The negative sign indicates that the velocity is directed downward.
Finally, to find the speed at impact, we can use the Pythagorean theorem:
speed = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
speed = sqrt((1.4 m/s)^2 + (-4.2 m/s)^2)
speed ≈ 4.4 m/s (rounded to one decimal place)
The speed of the mug at impact is approximately 4.4 m/s.
For the direction, it is mentioned in the question that the direction is "below the horizontal." This means that the mug is moving downwards at an angle with respect to the horizontal. Therefore, the direction of the mug at impact is vertically downward.