Consider a gun fired from a tower 96ft above the ground. The angle of elevation of the gun is 60 degrees, and its muzzle velocity is 1600 ft/sec (also gravity in this equation is 32 ft/sec^2).
a)When does the bullet hit the ground?
b)Write a parametric equation that models the path of the bullet.
c)What is the range of the gun?
d) find the equation for the bullet based on rectangular coordinates. What type of curve is this?
x = u t
y = yi + vi t - 16 t^2
----------------------------
u = 1600 cos 60 = 800 ft/sec the whole time
so
x = u t
vi = 1600 sin 60 = 1386 ft/s
so
y = 96 + 1386 t -16 t^2
so
when is y = 0 ?
0 = 96 + 1386 t - 16 t^2
16 t^2 - 1386 t - 96 = 0
so
t = 86.7 or -.07 (forget negative root)
so hits ground in 86.7 seconds
do u t for range
for part d use t = x/u, u is constant
y = yi + vi (x/u) - 16 x^2/u^2
or
y = c + b x + a x^2 a parabola
To answer these questions, we can use principles of projectile motion and trigonometry. Let's break down the steps to find the solutions:
a) When does the bullet hit the ground?
First, we need to find the time it takes for the bullet to reach the ground. We can use the formula for vertical displacement:
d = v0 * t + (1/2) * a * t^2
Where:
d = vertical displacement (in this case, -96 ft since it is below the tower)
v0 = initial vertical velocity (which is the vertical component of the muzzle velocity, v_y0 = v0 * sin(angle))
a = acceleration due to gravity (-32 ft/sec^2)
t = time
Rearranging the equation, we get:
96 = v_y0 * t + (1/2) * (-32) * t^2
0 = -16 * t^2 + v_y0 * t - 96
Now, we can solve this quadratic equation for t using the quadratic formula:
t = (-v_y0 ± sqrt(v_y0^2 - 4 * (-16) * (-96))) / (2 * (-16))
Substituting the values, we have:
t = (-v_y0 ± sqrt(v_y0^2 + 12288)) / (-32)
Calculate the values of v_y0:
v_y0 = 1600 ft/sec * sin(60°) = 1385.64 ft/sec
Plug in the value of v_y0 into the equation:
t = (-1385.64 ± sqrt(1385.64^2 + 12288)) / (-32)
Calculating the two possible values of t, we find:
t1 ≈ 0.286 seconds
t2 ≈ 2.84 seconds
Since the bullet must hit the ground, we use t2 as the answer: the bullet hits the ground approximately 2.84 seconds after being fired.
b) Write a parametric equation that models the path of the bullet.
In a projectile motion, we can describe the bullet's motion using parametric equations:
x = v_x0 * t
y = v_y0 * t + (1/2) * a * t^2
Where:
x = horizontal displacement
y = vertical displacement
v_x0 = initial horizontal velocity (v_x0 = v0 * cos(angle))
v_y0 = initial vertical velocity (calculated as mentioned earlier)
a = acceleration due to gravity (-32 ft/sec^2)
t = time
Substituting the values, our parametric equations become:
x = (1600 ft/sec * cos(60°)) * t
y = (1385.64 ft/sec) * t + (1/2) * (-32 ft/sec^2) * t^2
Simplifying:
x = 800 ft/sec * t
y = 1385.64 ft/sec * t - 16 ft/sec^2 * t^2
c) What is the range of the gun?
The range of the gun refers to the horizontal distance the bullet travels before hitting the ground. We can find the range by determining the value of x when y = 0 (ground level).
Set y = 0 in the parametric equation for y:
1385.64 ft/sec * t - 16 ft/sec^2 * t^2 = 0
1385.64 t - 16 t^2 = 0
t(1385.64 - 16t) = 0
Solving this equation, we find two possible solutions:
t1 = 0 (initial time, before the bullet was fired)
t2 ≈ 86.60 seconds
Since t2 represents the time when the bullet hits the ground, we discard t1. Therefore, at t ≈ 86.60 seconds, the bullet has hit the ground.
Now, we can find the range (horizontal distance) by substituting t2 into the equation for x:
x = 800 ft/sec * t2
x ≈ 800 ft/sec * 86.60 seconds
x ≈ 69,281.08 ft
Therefore, the range of the gun is approximately 69,281.08 ft.
d) Find the equation for the bullet based on rectangular coordinates. What type of curve is this?
To find the equation in rectangular (Cartesian) coordinates, we need to eliminate the variable t from the parametric equations.
From the parametric equations:
x = 800 ft/sec * t
y = 1385.64 ft/sec * t - 16 ft/sec^2 * t^2
We can solve the first equation for t:
t = x / 800 ft/sec
Substituting t into the second equation:
y = 1385.64 ft/sec * (x / 800 ft/sec) - 16 ft/sec^2 * (x / 800 ft/sec)^2
Simplifying:
y = 1.73205x - 0.00213333x^2
Therefore, the equation for the bullet in rectangular coordinates is y = 1.73205x - 0.00213333x^2.
This equation represents a downward-opening parabolic curve, which is commonly observed in projectile motion.