Muzzle Velocity = initial velocity of bullet = [tex]V_i[/tex] = 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 = ?
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,
[tex]H = V_{iy}t + \frac{1}{2}gt^2[/tex]
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.
[tex]H = \frac{1}{2}gt^2[/tex]
[tex]H = \frac{1}{2} x 10 x (0.067)^2\\\\H = (5)x(4.489x10^{-3})\\\\H = 0.022445 meters[/tex]
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,
[tex]H = \frac{1}{2}gt^2\\\\H = \frac{1}{2}x10x(0.133)^2\\\\H = 0.088445 meters[/tex]
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,
[tex]H = \frac{1}{2}gt^2\\\\H = \frac{1}{2}x10x(0.2)^2 \\\\H = 0.2 meters[/tex]
D) By the second equation of motion,
[tex]H=\frac{1}{2}gt^2 \\\\H = \frac{1}{2}x10x(1)^2 \\\\H = 5 meters[/tex]
Angle = θ(theta) = 30
Initial velocity = [tex]V_0[/tex] = 200m/s
Time to reach the ground = T = ?
Range = R = ?
The total time of flight is given by,
[tex]T = \frac{2V_0Sin(theta)}{g}\\\\T = \frac{2(200)Sin(30)}{10} \\\\T = \frac{200}{10}\\\\T = 20 seconds[/tex]
the Range of projectile is given by,
[tex]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[/tex]
