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3 years ago

I need help with these questions :(see image )

Physics

2 answers:

vovangra [49]3 years ago

5 0

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 11)</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h3>DATA:</h3>

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 = ?

<h2>_____________________________________</h2><h3>SOLUTION:</h3>

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,

$H = V_{iy}t + \frac{1}{2}gt^2$

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.

$H = \frac{1}{2}gt^2$

$H = \frac{1}{2} x 10 x (0.067)^2\\\\H = (5)x(4.489x10^{-3})\\\\H = 0.022445 meters$

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,

$H = \frac{1}{2}gt^2\\\\H = \frac{1}{2}x10x(0.133)^2\\\\H = 0.088445 meters$

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,

$H = \frac{1}{2}gt^2\\\\H = \frac{1}{2}x10x(0.2)^2 \\\\H = 0.2 meters$

D) By the second equation of motion,

$H=\frac{1}{2}gt^2 \\\\H = \frac{1}{2}x10x(1)^2 \\\\H = 5 meters$

<h2>______________________________________Question 12:</h2><h3>DATA: </h3>

Angle = θ(theta) = 30

Initial velocity = $V_0$ = 200m/s

Time to reach the ground = T = ?

Range = R = ?

<h2>_____________________________________</h2><h3>SOLUTION: </h3>

The total time of flight is given by,

$T = \frac{2V_0Sin(theta)}{g}\\\\T = \frac{2(200)Sin(30)}{10} \\\\T = \frac{200}{10}\\\\T = 20 seconds$

the Range of projectile is given by,

$R = \frac{V_0^2}{g}Sin2(theta)\\\\R = \frac{200^2}{10} Sin2(30)\\\\R = \frac{40000}{10}Sin60\\\\R = (4000)x(0.866)\\\\R = 3464.1 meters$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

IRISSAK [1]3 years ago

4 0

Answer:number one is

Explanation:

I took this test

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