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tangare [24]
3 years ago
7

Suppose Tom Harmon17 is standing at the exact center of the Ohio State football field18 on the 50 yard line. The field is 300 fe

et long and 160 feet wide. Suppose the ball leaves Harmon’s hand at a height of six feet and has initial velocity v(0) = 40i + 39j + 32k feet per second where j points toward the sideline, i points toward the end zone, and k = i × j points straight up. The acceleration due to gravity is −32k feet per second squared and the mass of the football is .45 kilograms.
(a) Determine a parameterization for the position of the football at time t seconds after it is thrown.
(b) Suppose David Nelson19 catches the ball when it is six feet above the ground. Is Nelson in bounds or out of bounds when he catches the ball? (Assume Nelson’s toes are touching the ground directly below the ball when he catches the ball.)'
Physics
1 answer:
nignag [31]3 years ago
6 0

Answer:

a) x = 40 t , y = 39 t ,  z = 6 + 32 t - 16 t ²,   b)     x = 80 feet ,  y = 78 feet , the ball came into the field  

Explanation:

a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis

Since the cast is in the center of the field, let's place the coordinate system

          x₀ = 0

          y₀ = 0

          z₀ = 6 feet

x-axis (towards end zone,  GOAL zone)

         x = xo + v₀ₓ t

        x = 40 t

y-axis (field width)

        y = y₀ + v_{oy} t

        y = 39 t

z axis (vertical)

        z = z₀ + v_{oz} t - ½ g t²

        z = 6 + 32 t - ½ 32 t²

        z = 6 + 32 t - 16 t ²

b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive

          z = 6

          6 = 6 + 32 t - 16 t²

          (t - 2)t  = 0

           t=0 s

           t= 2 s

The ball position

          x = 40 2

          x = 80 feet

           

          y = 39 2

          y = 78 feet

the dimensions of the field from the coordinate system (center of the field) are

             x_total = 150 feet

             y _total = 80 feet

so we can see that the ball came into the field

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4. The Mariana trench is in the Pacific Ocean and has a depth of approximately 11,000 m. The density of seawater is approximatel
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Explanation:

It is known that relation between pressure and density is as follows.

            P = \rho gh

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            h = height

Putting the given values into the above formula as follows.

              P = \rho gh

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Now, relation between pressure and force is as follows.

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or,            F = PA

                F = 110495000 \times \pi \times (0.1)^{2}

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The perceived loudness of a sound is measured in
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5 0
3 years ago
Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig
MissTica

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

         Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

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6 0
3 years ago
In a race, Usain Bolt accelerates at
jeka94

Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

3 0
3 years ago
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