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____ [38]
3 years ago
11

During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 1.4

km/s at an initial inclination of 65.8 ◦ to the horizontal. The acceleration of gravity is 9.8 m/s 2 . How far away did the shell hit?
Physics
1 answer:
Ksju [112]3 years ago
5 0

Answer:

The shell hit at a horizontal distance 149.41 km

Explanation:

The shell had an initial speed of 1.4 km/s

Initial speed = 1400 m/s

Inclination = 65.8° to the horizontal

First let us find the time of travel

Consider the vertical motion,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 1400 sin 65.8 = 1276.97 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = 0 m

Substituting

                 s = ut + 0.5 at²          

                 0 = 1276.97 x t + 0.5 x  (-9.81) xt²

                  t = 260.34 s

Now let us find the horizontal displacement,

Consider the horizontal motion,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 1400 cos 65.8 = 573.89 m/s

Acceleration, a = 0 m/s²

Time, t = 260.34 s

Substituting

                 s = ut + 0.5 at²          

                 s = 573.89 x 260.34 + 0.5 x  0 x 260.34²

                  s = 149407.11 m = 149.41 km

The shell hit at a horizontal distance 149.41 km

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3 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

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We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

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We will calculate the time of flight,

   Y = Voy t – ½ g t2  

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When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

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0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

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The value of zero corresponds to the departure point and the flight time is 15.3 s

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R2 = 25² sin (2 30) / 9.8

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R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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3 years ago
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