A projectile is fired with an initial velocity of v0 feet per second. The projectile can be pictured as being fired from the origin into the first quadrant, making an angle O with the positive x-axis. If there is no air resistance, then at t seconds the coordinates of the projectile are x=v0(t)(cos O) and y=-16t^2+v0(t)(sin O). suppose a projectile leaves the gun at 100 ft/sec and O=60 degrees.

A) what are the coordinates of the projectile at time t=4 sec?
b)For how many seconds is the projectile in the air?
c)How far fro the gun does the projectile land?
d)What is the max height attained?
e)find an expression in terms of v0 and O for the time in the air
f)find an expression in terms of v0 and O for the distance from the gun
g) Find an expression in terms of v0 and O for the maximum height
h) show that y= ((-16sec^2O)/(v0)^2)(x^2)+xtanO

Vo = 100Ft/s[60o]

X = 4*100*cos60 = 200 Ft.
Y = 4*100*sin60 = 346.4 Ft.

a. (X,Y) = (200,346.4).

b. Yo = Vo*sin O = 100^sin60 = 86.6 Ft/s

Y = Yo + g*t = 0 @ max. Ht.
86.6 + (-32)t = 0
-32t = -86.6
Tr = 2.71 s. = Rise time.

Tf = Tr = 2.71 s. = Fall time.

T = Tr + Tf = 2.71 + 2.71 = 5.42 s. =
Time in air.

c. D = Vo^2*sin(2*O)/g
D = 100^2*sin(120)/64 = 271 Ft.

d. Y^2 = Yo^2 + 2g*h
h = (Y^2-Yo^2)/2g
h = (0-86.6^2)/-64 = 117 Ft.

e. T = -2*Vo*sin(O)/g

Xo* T = 100*cos60 * 17.7=884 Ft.

d.

To solve the given problem, we need to use the equations of motion for projectile motion. Let's go through each part step by step.

A) To find the coordinates of the projectile at time t=4 sec, we can use the equations:
x = v0(t)(cos O)
y = -16t^2 + v0(t)(sin O)

Substituting the given values, v0 = 100 ft/sec and O = 60 degrees, we have:
x = 100(4)(cos 60) = 200 ft
y = -16(4)^2 + 100(4)(sin 60) = 96 ft

So, the coordinates of the projectile at time t=4 sec are (200 ft, 96 ft).

B) The projectile is in the air as long as y > 0. We can set y = 0 and solve for the time t:
-16t^2 + 100t(sin 60) = 0
-16t(t - 5) = 0

This equation gives us two possible values for t: t = 0 (at the start) and t = 5 sec (landing time). Therefore, the projectile is in the air for 5 - 0 = 5 seconds.

C) To find how far the projectile lands from the gun, we need to find the final x-coordinate when y = 0. Setting y = 0 and solving for x:
-16t^2 + 100t(sin 60) = 0
-16t(t - 5) = 0

Again, this equation gives us two possible values for t: t = 0 and t = 5 sec. At t = 0, the projectile is at the starting point (x = 0). At t = 5, the projectile lands. Therefore, the projectile lands at x = 100(5)(cos 60) = 250 ft from the gun.

D) The maximum height attained can be found by finding the vertex of the parabolic equation y = -16t^2 + 100t(sin 60). The vertex occurs at the time t = -b/(2a), where a = -16 and b = 100(sin 60).
t = -100(sin 60)/(2(-16)) = 1.25 sec

Substituting t = 1.25 sec into the equation, we get:
y = -16(1.25)^2 + 100(1.25)(sin 60) = 106.25 ft

So, the maximum height attained is 106.25 ft.

E) The time in the air, T, can be expressed in terms of v0 and O.
Using the equation y = -16t^2 + v0t(sin O) and setting y = 0, we have:
-16t^2 + v0t(sin O) = 0
t(t - v0(sin O)/16) = 0

This equation gives us two possible values for t: t = 0 and t = v0(sin O)/16. Therefore, the time in the air is T = v0(sin O)/16 seconds.

F) The distance from the gun can be expressed in terms of v0 and O.
Using the equation x = v0t(cos O) and substituting T = v0(sin O)/16, we have:
x = v0(T)(cos O) = v0(v0(sin O)/16)(cos O)
x = (v0^2/16)(sin O)(cos O)

G) The maximum height can be expressed in terms of v0 and O.
Using the equation y = -16t^2 + v0t(sin O) and substituting the time t = 1.25 sec, we have:
y = -16(1.25)^2 + v0(1.25)(sin O)
y = -16(1.5625) + v0(1.25)(sin O)
y = -25 + (5/4)v0(sin O)

H) To show that y = ((-16sec^2O)/(v0)^2)(x^2) + xtanO, we can use the trigonometric identities:
sin^2 O + cos^2 O = 1 (1)
tan O = sin O / cos O (2)

Starting from the given equation y = -16t^2 + v0t(sin O) and substituting t = x / (v0(cos O)), we have:
y = -16(x / (v0(cos O)))^2 + v0(x / (v0(cos O)))(sin O)
y = -16(x^2 / v0^2(cos O)^2) + (x / cos O)(sin O)
y = (-16/cos^2 O)(x^2 / v0^2) + (sin O / cos O)(x)
y = (-16sec^2 O)(x^2 / v0^2) + xtan O

So, we have shown that y = ((-16sec^2 O)/(v0^2))(x^2) + xtan O.