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

See previous post: Sun,11:33 AM.

To answer the questions, we will use the given formulas for the x and y coordinates of the projectile:

x = v0(t) * cos(O)
y = -16t^2 + v0(t) * sin(O)

Let's proceed with each question step by step:

a) To find the coordinates of the projectile at t = 4 seconds, we substitute t = 4 into the equations:

x = v0(4) * cos(60)
= 100(4) * 0.5
= 200 feet

y = -16(4)^2 + 100(4) * sin(60)
= -256 + 400 * 0.866
= 210.5 feet

So, at t = 4 seconds, the coordinates of the projectile are (200, 210.5).

b) The projectile is in the air until its y-coordinate becomes 0. Set y = 0 and solve for t:

-16t^2 + 100t * sin(60) = 0
-16t^2 + 50t = 0
t(-16t + 50) = 0

This equation has two solutions: t = 0 and t = 50/16. Since t = 0 represents when the projectile is launched and not in the air, we consider the positive solution:

t = 50/16 ≈ 3.125 seconds

Therefore, the projectile is in the air for approximately 3.125 seconds.

c) To find the distance from the gun where the projectile lands, we need to find the x-coordinate when y = 0. Set y = 0 and solve for x:

-16t^2 + v0(t) * sin(O) = 0
-16(t^2) + 100(t) * sin(60) = 0
-16(t^2) + 50(t) = 0
t(-16t + 50) = 0

Again, we have two solutions: t = 0 and t = 50/16. Considering the positive solution:

t = 50/16 ≈ 3.125 seconds

Now we substitute this value of t into the x-coordinate formula:

x = v0(t) * cos(O)
= 100(3.125) * cos(60)
= 156.25 feet

Therefore, the projectile lands approximately 156.25 feet from the gun.

d) The maximum height of the projectile is attained at the vertex of the parabolic trajectory, which occurs when the derivative of y with respect to t equals zero. We find the derivative of y with respect to t:

dy/dt = -32t + v0 * sin(O)

Setting dy/dt = 0, we have:

-32t + 100 * sin(60) = 0
-32t + 100 * √3/2 = 0
-32t + 50√3 = 0
32t = 50√3
t = (50√3)/32

Substituting this value of t into the y-coordinate formula:

y = -16t^2 + v0(t) * sin(O)
= -16[(50√3)/32]^2 + 100 * (50√3/32) * sin(60)
≈ 42.85 feet

So, the maximum height attained by the projectile is approximately 42.85 feet.

e) The time in the air can be expressed in terms of v0 and O by considering the y-coordinate. Since the projectile lands when y = 0, we can solve the equation for t:

-16t^2 + v0(t) * sin(O) = 0
-16t^2 + 100(t) * sin(O) = 0
t(-16t + 100sin(O)) = 0

Since t cannot be zero (as it represents the launch), we have:

-16t + 100sin(O) = 0
16t = 100sin(O)
t = 100sin(O)/16
t = 25sin(O)/4

Therefore, the time in the air can be expressed as t = (25sin(O))/(4v0).

f) The distance from the gun can be expressed in terms of v0 and O by considering the x-coordinate. When y = 0, we can solve the equation for x:

x = v0(t) * cos(O)
= 100(25sin(O)/4) * cos(O)
= 25(100/4)(sin(O))(cos(O))
= 625sin(O)cos(O)

Therefore, the distance from the gun can be expressed as x = 625sin(O)cos(O).

g) The maximum height can be expressed in terms of v0 and O by substituting the obtained value of t into the y-coordinate formula:

y = -16t^2 + v0(t) * sin(O)
= -16[(25sin(O)/4)^2] + 100(25sin(O)/4) * sin(O)
= -16[625sin^2(O)/16] + 100(25sin(O)/4)sin(O)
= -625sin^2(O) + 625sin^2(O)
= 625sin^2(O)

Therefore, the maximum height can be expressed as y = 625sin^2(O).

h) To show that y = ((-16sec^2(O))/(v0^2))(x^2) + xtan(O), we can substitute the given formulas for x and y:

x = v0(t) * cos(O) = 100t * cos(O)
y = -16t^2 + v0(t) * sin(O) = -16t^2 + 100t * sin(O)

First, rearrange the x formula to get t in terms of x:

t = x / (100cos(O))

Now, substitute this value of t into the y formula:

y = -16(x^2 / [100^2cos^2(O)]) + 100(x / 100cos(O)) * sin(O)
= -16x^2 / 100cos^2(O) + xsin(O) / cos(O)
= -16x^2 / 100cos^2(O) + xsin(O)cos(O) / cos(O)^2
= (-16x^2 + xtan(O)cos(O)) / [100cos^2(O)]
= ((-16x^2 + xtan(O)cos(O))/(v0^2))(100/v0^2)
= ((-16x^2)/(v0^2) + (xtan(O)cos(O))/(v0^2))(100/v0^2)

Simplifying the expression, we get:

y = ((-16sec^2(O))/(v0^2))(x^2) + xtan(O)

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