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marysya [2.9K]
3 years ago
6

A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table

?
Question 3 options:

A 3.3 m

B 1.1 m

C 2.2 m

D 3.0 m
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3


consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m


C 2.2 m

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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
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b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

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   The angular speed at the lowest point is w = 1.09 \ rad/s

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               I = \frac{(0.140) (1.4)^2}{3}

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Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

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   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

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