Question

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

Answer 1

Hey There!

_____________________________________

Question 11)

_____________________________________

Answer:

_____________________________________

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 = ?

_____________________________________

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,

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

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,

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

______________________________________Question 12:

DATA:

Angle = θ(theta) = 30

Initial velocity = $V_0$ = 200m/s

Time to reach the ground = T = ?

Range = R = ?

_____________________________________

SOLUTION:

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,

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

_____________________________________

Best Regards,

'Borz'

Answer 2

Answer:number one is

a

Explanation:

I took this test