A pistol that fires a signal flare gives the flare an initial speed of 180m/s. Air resistance is ignored.
(a)If the flare is fired at an angle of 55° above the horizontal, what is the horizontal range?
(b)If the flare is fired at the same angle over the flat of sea tranquility on the moon where g=1.6m/s2, what is the horizontal range?
Pls help
Vo = 180m/s[55o].
a. Range = Vo^2*sin(2A)/g = 180^2*sin(110)/9.8 = 3107 m.
b. Same procedure as "a" except g = 1.6 m/s^2.
To find the horizontal range of the flare, we can break the initial velocity into its horizontal and vertical components.
(a) On Earth:
The horizontal component of the velocity is given by Vx = V * cos(theta), where V is the initial speed of the flare and theta is the angle of projection.
Vx = 180 m/s * cos(55°)
Vx ≈ 180 m/s * 0.5736
Vx ≈ 103.44 m/s
The horizontal range (R) can be determined using the formula:
R = (Vx^2 * sin(2 * theta)) / g, where g is the acceleration due to gravity (9.8 m/s^2).
R = (103.44 m/s)^2 * sin(2 * 55°) / 9.8 m/s^2
R ≈ 14482.782 m
Round to: R ≈ 14480 m
Therefore, the horizontal range of the flare fired on Earth is approximately 14480 meters.
(b) On the Moon:
Since gravity on the Moon is weaker, we need to adjust for the acceleration due to gravity.
g_moon = 1.6 m/s^2
Using the same formula as before, the horizontal range on the Moon can be calculated as follows:
R_moon = (Vx^2 * sin(2 * theta)) / g_moon
R_moon = (103.44 m/s)^2 * sin(2 * 55°) / 1.6 m/s^2
R_moon ≈ 4045.5498 m
Round to: R_moon ≈ 4046 m
Therefore, the horizontal range of the flare fired on the Moon is approximately 4046 meters.
To calculate the horizontal range of the flare in both scenarios, we can use the equations of projectile motion.
(a) Calculating the horizontal range on Earth:
Let's break down the initial velocity into its horizontal and vertical components. The vertical component is given by Vfy = V * sin(θ), and the horizontal component is Vfx = V * cos(θ), where V is the initial speed of 180 m/s and θ is the firing angle of 55°.
The vertical component determines the time of flight, which can be calculated using the equation:
t = (2 * Vfy) / g
where g is the acceleration due to gravity, approximately 9.8 m/s² on Earth.
Now, we can calculate the horizontal range using the formula:
R = Vfx * t
Substituting the values, we get:
Vfy = 180 m/s * sin(55°)
t = (2 * Vfy) / g
R = Vfx * t
(b) Calculating the horizontal range on the Moon:
In this scenario, the acceleration due to gravity is different, equal to 1.6 m/s².
Using the same steps as in part (a), we can calculate the horizontal range. The only difference is that we would use the Moon's acceleration due to gravity, g = 1.6 m/s², instead of Earth's g.
Substituting the values into the equations, we can calculate the horizontal range.
In both cases, it is important to convert the firing angle from degrees to radians if needed.
I hope this helps you calculate the horizontal range of the flare in both scenarios! Let me know if you need any further assistance.