Question

I need help with these questions :
(see image )​

Answer 1

Answer:number one is

a

Step-by-step explanation:

I took this test

Answer 2

Hey There!

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Question 11)

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Answer:

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DATA:

Muzzle Velocity = initial velocity of bullet =
`V_i` = 300 m/s

Height "H" dropped at horizontal distance,

a) x = 20m

b) x = 40m

c) x = 60m

d) How far will it drop after 1 second = ?

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SOLUTION:

We have to find the Height H dropped at the respective given horizontal distances. To find H first we have to calculate the time the bullet has taken to reach at every "X"(Horizontal Distance). So,

A) X = 20m

v = 300m/s

t = ?

X = vt

20 = 300 x t

t = 0.067 seconds.

Now, Height dropped can be determine by 2nd equation of motion,

`H = V_(iy)t + (1)/(2)gt^2`

As initial vertical velocity is zero because the question says that the bullet is fired in the horizontal direction and there is no velocity component in y-axis or vertical axis.

`H = (1)/(2)gt^2`

`H = (1)/(2) x 10 x (0.067)^2\\\\H = (5)x(4.489x10^(-3))\\\\H = 0.022445 meters`

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B) X = 40m

v = 300m/s

t = ?

X = vt

40 = 300 x t

t = 0.133 seconds

Now, Height dropped can be determine by 2nd equation of motion,

`H = (1)/(2)gt^2\\\\H = (1)/(2)x10x(0.133)^2\\\\H = 0.088445 meters`

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C) X = 60m

v = 300m/s

t = ?

X = vt

60 = 300 x t

t = 0.2 seconds

Now, Height dropped can be determine by 2nd equation of motion,

`H = (1)/(2)gt^2\\\\H = (1)/(2)x10x(0.2)^2 \\\\H = 0.2 meters`

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D) By the second equation of motion,

`H=(1)/(2)gt^2 \\\\H = (1)/(2)x10x(1)^2 \\\\H = 5 meters`

______________________________________Question 12:

DATA:

Angle = θ(theta) = 30

Initial velocity =
`V_0` = 200m/s

Time to reach the ground = T = ?

Range = R = ?

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SOLUTION:

The total time of flight is given by,

`T = (2V_0Sin(theta))/(g)\\\\T = (2(200)Sin(30))/(10) \\\\T = (200)/(10)\\\\T = 20 seconds`

the Range of projectile is given by,

`R = (V_0^2)/(g)Sin2(theta)\\\\R = (200^2)/(10) Sin2(30)\\\\R = (40000)/(10)Sin60\\\\R = (4000)x(0.866)\\\\R = 3464.1 meters`

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Best Regards,

'Borz'