Answer:
Explanation:
To determine the angle of elevation needed to hit an object 160 meters away with a muzzle speed of 60 meters per second, we can use the kinematic equations and principles of projectile motion.
Given:
Muzzle speed (initial velocity), u = 60 m/s
Distance to the object, x = 160 m
Acceleration due to gravity, g = 9.8 m/s^2
To Find:
Angle of elevation, θ
Solution:
We can break the initial velocity into horizontal and vertical components. The horizontal component remains constant throughout the projectile's motion, while the vertical component is affected by gravity.
Horizontal component: u_x = u * cos(θ)
Vertical component: u_y = u * sin(θ)
Using the horizontal motion equation, we can find the time of flight, t:
x = u_x * t
t = x / u_x
Using the vertical motion equation, we can find the time it takes to reach the maximum height, t_max:
u_y = u * sin(θ)
v_y = 0 at the maximum height
v_y = u_y - g * t_max
0 = u * sin(θ) - g * t_max
t_max = u * sin(θ) / g
The total time of flight is given by:
t_total = 2 * t_max
Since the distance traveled horizontally is equal to x, we can rewrite the horizontal motion equation as:
x = u_x * t_total
x = u * cos(θ) * t_total
Substituting the values and equations, we have:
160 = 60 * cos(θ) * (2 * (u * sin(θ) / g))
160 = 120 * (sin(θ) / g) * cos(θ)
160 = (120 / g) * sin(θ) * cos(θ)
Simplifying the equation further, we can use the identity sin(2θ) = 2 * sin(θ) * cos(θ):
160 = (120 / g) * (1/2) * sin(2θ)
Rearranging the equation to isolate sin(2θ), we get:
sin(2θ) = (160 * 2 * g) / (120)
Taking the inverse sine (arcsin) of both sides to find 2θ, we have:
2θ = arcsin((160 * 2 * g) / (120))
Finally, we can find θ by dividing the result by 2:
θ = arcsin((160 * 2 * g) / (120)) / 2
Using the given value of g = 9.8 m/s^2, we can substitute it into the equation and calculate θ.
Final Answer:
The angle of elevation needed to hit an object 160 meters away with a muzzle speed of 60 meters per second, neglecting air resistance, is given by θ = arcsin((160 * 2 * 9.8) / 120) / 2.