Answer :
Answer:
a) The bullet hits the ground after 64 seconds.
b) The bullet hits the ground 113,511.7 feet away.
c) The maximum height attained by the bullet is of 16,384 feet.
Step-by-step explanation:
Equations of motion:
The equations of motion for the bullet are:
[tex]x(t) = (v_0\cos{\alpha})t[/tex]
[tex]y(t) = (v_0\sin{\alpha})t - 16t^2[/tex]
In which [tex]v_0[/tex] is the initial speed and [tex]\alpha[/tex] is the angle.
Initial speed of 2048 ft/s at an angle of 30o to the horizontal.
This means that [tex]v_0 = 2048, \alpha = 30[/tex].
So
[tex]x(t) = (v_0\cos{\alpha})t = (2048\cos{30})t = 1773.62t[/tex]
[tex]y(t) = (v_0\sin{\alpha})t - 16t^2 = (2048\sin{30})t - 16t^2 = 1024t - 16t^2[/tex]
(a) After how many seconds will the bullet hit the ground?
It hits the ground when [tex]y(t) = 0[/tex]. So
[tex]1024t - 16t^2 = 0[/tex]
[tex]16t^2 - 1024t = 0[/tex]
[tex]16t(t - 64) = 0[/tex]
16t = 0 -> t = 0 or t - 64 = 0 -> t = 64
The bullet hits the ground after 64 seconds.
(b) How far from the gun will the bullet hit the ground?
This is the horizontal distance, that is, the x value, x(64).
[tex]x(64) = 1773.62(64) = 113511.7[/tex]
The bullet hits the ground 113,511.7 feet away.
(c) What is the maximum height attained by the bullet?
This is the value of y when it's derivative is 0.
We have that:
[tex]y^{\prime}(t) = 1024 - 32t[/tex]
[tex]1024 - 32t = 0[/tex]
[tex]32t = 1024[/tex]
[tex]t = \frac{1024}{32} = 32[/tex]
At this instant, the height is:
[tex]y(32) = 1024(32) - 16(32)^2 = 16384[/tex]
The maximum height attained by the bullet is of 16,384 feet.