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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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Answer:

12.3 seconds

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To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r
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Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

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a).

ρ= 1.2 \frac{kg}{m^{3} }, A_{t}= 0.7 m^{2}, D_{t}= 0.28

F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N

b).

ρ= 1.2 \frac{kg}{m^{3} }, A_{h}= 2.44 m^{2}, D_{h}= 0.57

F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N

6 0
3 years ago
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3 0
3 years ago
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scoundrel [369]

Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original  so it has 8.8 cm with the same orientation as original  and it is a virtual imagen.

Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:

1/p+1/q=1/f  where p and q represents the distance to the mirror  for the object and imagen, respectively. f is the focal length for the concave mirror.

replacing the values we obtain:

1/8.3+1/q=1/15.2

so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3

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The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2

We also add a picture to see the imagen formation for this case.

6 0
3 years ago
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