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spayn [35]
3 years ago
10

A ball is thrown upward from the top of a 25.0 m tall building. The ball’s initial speed is 12.0 m/sec. At the same instant, a p

erson is running on the ground at a distance of 31.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?
Physics
1 answer:
zimovet [89]3 years ago
6 0
<h2>Person must have 8.18 m/s to catch the ball</h2>

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 12 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -25 m

Substituting

             -25 = 12 x t + 0.5 x -9.81 x t²

               4.905 t² -12t - 25 = 0

              t = 3.79 sec

Ball hits ground after 3.79 seconds.

So person need to cover 31 m in 3.79 seconds

Consider the horizontal motion of person

We have equation of motion s = ut + 0.5at²

Initial velocity, u = ?

Acceleration, a = 0 m/s²

Displacement, s = 31 m

Time, t = 3.79 seconds

Substituting

             31 = u x 3.79 + 0.5 x 0 x 3.71²

               u = 8.18 m/s

Person must have 8.18 m/s to catch the ball

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3 years ago
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Which has a total mass of 1612 kg. If she accelerates from rest to a speed of 12.87 m/s in 3.47 s, what is the minimum power req
blondinia [14]

Answer:

76969.29 W

Explanation:

Applying,

P = F×v............. Equation 1

Where P = Power, F = force, v = velocity

But,

F = ma.......... Equation 2

Where m = mass, a = acceleration

Also,

a = (v-u)/t......... Equation 3

Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s

Substitute these values into equation 3

a = (12.87-0)/3.47

a = 3.71 m/s²

Also Given: m = 1612 kg

Substitute into equation 2

F = 1612(3.71)

F = 5980.52 N.

Finally,

Substitute into equation 1

P = 5980.52×12.87

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2 years ago
WHO WANTS BRAINLIEST THEN ANSWER THIS QUESTION
lozanna [386]

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6 0
3 years ago
To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

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