Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.15 m. the mug slides off the counter and strikes the floor 1.50 m from the base of the counter. (a) with what velocity did the mug leave the counter? (b) what was the direction of the mug's velocity just before it hit the floor?

Answer 1

A.    Let’s say the horizontal component of the velocity is vx and the vertical is vy. 
Initially at t=0 (as the mug leaves the counter) the components are v0x and v0y. 
v0y = 0 since the customer slides it horizontally so applied force is in the x component only.

The equations for horizontal and vertical projectile motion are:
x = x0 + v0x t 
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 

Setting the origin to be the end corner of the counter so that x0=0 and y0=0, hence:

x = v0x t

y = - 1/2 g t^2 

Given value are: x=1.50m and y=-1.15m (y is negative since mug is going down)

1.50m = v0x t    ---->  v0x= 1.50/t

-1.15m = -(1/2) (9.81) t^2    -----> t =0.4842 s

Calculating for v0x:

v0x = 3.10 m/s

B.    v0x is constant since there are no other horizontal forces so, v0x=vx=3.10m/s

vy can be calculated from the formula:

vy = v0y + at         where a=-g (negative since going down)

vy = -gt = -9.81 (0.4842)

vy = -4.75 m/s

 

Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy

tan(90-Ø )= 3.1/4.75

Ø = 56.87˚ below the horizontal

Answer 2

(a) The velocity of mug was 3.10 m/s

(b) The direction of the mug's velocity just about it hit the floor is about 56.9° below the horizontal

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Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

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This problem is about Projectile Motion.

Given:

horizontal distance = x = 1.50 m

height of the counter = h = 1.15 m

Unknown:

a. initial velocity = u = ?

b. direction of the mug's velocity = θ = ?

Solution:

Question a:

$x = ut$

$1.5 = ut$

$t = \frac{1.5}{u}$ → Equation 1

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$h = \frac{1}{2}gt^2$

$h = \frac{1}{2}g(\frac{1.5}{u})^2$ ← Equation 1

$1.15 = \frac{1}{2}(9.8)(\frac{1.5}{u})^2$

$1.15 = \frac{441}{40} \div u^2$

$u^2 = \frac{441}{40} \div 1.15$

$u^2 = \frac{441}{46}$

$u = \sqrt{\frac{441}{46}}$

$u \approx 3.10 \texttt{ m/s}$

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Question b:

$\theta = \tan^{-1} \frac{v_y}{v_x}$

$\theta = \tan^{-1} \frac{-gt}{u}$

$\theta = \tan^{-1} \frac{-g\frac{x}{u}}{u}$

$\theta = \tan^{-1} \frac{-gx}{u^2}$

$\theta = \tan^{-1} \frac{-9.8(1.5)}{3.10^2}$

$\theta = \tan^{-1} (-\frac{23}{15})$

$\theta \approx -56.9^o \texttt{ to the horizontal}$

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Learn more

• Velocity of Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922
• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall