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6. A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle
of 30.0° from the horizontal with a velocity of 40.0 m/s. How far will it travel in the air?
7. A golfer is teeing off on a 170.0 m long par 3 hole. The ball leaves with a velocity of 40.0 m/s at
50.0° to the horizontal. Assuming that she hits the ball on a direct path to the hole, how far from
the hole will the ball land (no bounces or rolls)?
8. A punter in a football game kicks a ball from the goal line at 60.0° from the horizontal at 25.0
m/s.
a) What is the hang time of the punt (time in air)? (4.41 s)
b) How far down field does the ball land? (55.2 m)
9. A lad wants to throw a bag into the open window of his friend's room 10.0 m above. Assuming it
just reaches the window, he throws the bag at 60.0° to the ground:
a) At what velocity should he throw the bag? (16.2 m/s at 60.0° to the ground]
b) How far from the house is he standing when he throws the bag? (11.5 m]
10. An elastic loaded balloon launcher fires balloons at an angle of (38.0° N of E) from the surface
of the ground. If the initial velocity is 25.0 m/s, find how far away the balloons are from the
launcher when they hit the level ground again. [61.8 m]

Relax

Respuesta :

The kinematic relations we can find the results for the different short questions are:

6.  Range is    R = 141.4 m

7. Range is    R = 251.2 m

8.  a) Time  t = 4.42 s  b) Range  x = 55.25 m

9. a) Initial speed es v₀ = 16.16 m / s   b) range is  x = 11.55 m

10. Range  R = 6 1.9 m

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

6. They ask the range of the home run.  

Indicates initial velocity of 40.0 m / s and departure angle 30º

        R =[tex]\frac{v_o^2 sin 2 \theta }{g}[/tex]  

        R = [tex]\frac{40.0^2 sin\ 2 \ 30 }{9.8 }[/tex]  

        R = 141.4 m

7) Ask where the golf ball arrives to the ground.

They indicate the initial velocity of 40.0 m / s and the angle of 50º

        R = [tex]\frac{50.0^2 sin\ 2 \ 50 }{9.8}[/tex]

        R = 251.2 m

8) They indicate the initial velocity of 25.0 m / s and an angle of 60.0º

a) Time in the air.

When the ball reaches the ground its height is zero.

       y = [tex]v_{oy}[/tex]  t - ½ g t²

       0 = ([tex]v_{oy}[/tex]  - ½ g t) t

the solution is

      t=0               exit point

      t = [tex]\frac{2 v_{oy}}{ g}[/tex]  

vertical speed is

      [tex]v_{oy}[/tex] =  v₀ sin  60

Let's substitute

      t = [tex]\frac{2 \ 36 \ sin 60}{9.8}[/tex]  

      t = 4.42 s

b) How far does the ball go

      x = v₀ₓ t

      x = v₀ cos 60  t

      x = 25 4.42 cos 60

      x = 55.25 m

9) They indicate the height of the window y = 10.0 m and the launch angle is  60º

a) What is the initial speed of the throw

As it barely reaches the window, its vertical speed is zero

      [tex]v_y^2 = v_{oy}^2 - 2 g y \\0 = v_{o}^2 sin ^2 \theta - 2 g y\\ \\v_o^2 = \frac{2gy}{sin^2 \theta }[/tex]

      v₀² = [tex]\frac{2 \ 9.8 \ 10.0}{9.8}[/tex]

      v₀ = [tex]\sqrt{261.33}[/tex]  

      v₀ = 16.16 m / s

b) What is the horizontal distance

Let's find the time until we reach the window

       [tex]v_y = v_{oy} - gt\\0 = v_{oy} - gt[/tex]  

       t = [tex]\frac{v_o sin \theta}{g}[/tex]

       t = [tex]\frac{16.16 sin 60}{9.8}[/tex]

       t = 1.43 s

The horizontal distance is

     x = v₀ₓ t

     x = 1(6.16 cos 60) 1.43

     

10) The range of the balloons

      [tex]R = \frac{v_o^2 sin 2 \theta }{g}\\R = \frac{25.0^2 sin \ 2 \ 38 }{9.8}[/tex]  

      R = 6 1.9 m

In conclusion using the kinematics relations we can find the results for the different short questions are:

       6.  Range is    R = 141.4 m

       7. Range is    R = 251.2 m

       8.  a) Time  t = 4.42 s  b) Range  x = 55.25 m

       9. a) Initial speed es v₀ = 16.16 m / s   b) range is  x = 11.55 m

      10. Range  R = 6 1.9 m

Learn more here: brainly.com/question/10903823

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