A projectile is launched up and to the right over flat, level ground. Its range is equal to half of its maximum elevation above the ground. What was the angle between its initial velocity and the ground? Ignore air resistance.
° above the horizonontal
For projectile motion,
Range = R = (u^2)(sin2θ)/g
Maximum height = H = (u^2)(sinθ)^2/(2g)
As per the question, 2R = H
=> 2(u^2)(sin2θ)/g = (u^2)(sinθ)^2/(2g)
=> 4sin2θ = (sinθ)^2
=> 8sinθcosθ = (sinθ)^2
=> 8cosθ = sinθ
=> tanθ = (8)
=> θ = arctan(8)
To solve this problem, we can use the kinematic equations of projectile motion.
Let's assume that the initial velocity of the projectile is "v" and the angle between the initial velocity and the ground is "θ".
The range of a projectile, ignoring air resistance, is given by the equation:
R = (v^2 * sin(2θ))/g
where R is the range, v is the initial velocity, θ is the angle, and g is the acceleration due to gravity.
In this case, it is given that the range is equal to half of the maximum elevation above the ground. So, we have:
R = (1/2) * h
where h is the maximum elevation.
From the equation for the range, we can solve for the angle θ:
(1/2) * h = (v^2 * sin(2θ))/g
Rearranging the equation, we get:
sin(2θ) = (2 * R * g) / (v^2)
Taking the inverse sin (sin^(-1)) on both sides:
2θ = sin^(-1)((2 * R * g) / (v^2))
Dividing by 2, we get:
θ = (1/2) * sin^(-1)((2 * R * g) / (v^2))
Therefore, the angle between the initial velocity and the ground is θ = (1/2) * sin^(-1)((2 * R * g) / (v^2)) degrees.
Note: Make sure to input the corresponding units for the variables (e.g., range in meters, initial velocity in meters per second, etc.) in order to obtain the correct answer.
To find the angle between the initial velocity and the ground, we need to understand the relationship between the range (horizontal distance) and the maximum elevation (vertical distance) of the projectile.
Let's consider the motion of the projectile. Since there is no air resistance, the only force acting on the projectile is gravity. This means the projectile follows a parabolic trajectory.
Now, let's break down the motion into horizontal and vertical components. The horizontal component of the initial velocity remains constant throughout the motion, while the vertical component changes due to the effect of gravity.
Let's define some variables:
- v₀x is the initial velocity in the horizontal direction.
- v₀y is the initial velocity in the vertical direction.
- θ is the angle between the initial velocity and the ground.
The initial velocity can be divided into its horizontal and vertical components as follows:
v₀x = v₀ × cos(θ)
v₀y = v₀ × sin(θ)
Now, we know that the range (R) is equal to half of the maximum elevation (H):
R = 0.5 × H
The range can be expressed as the product of the horizontal component of velocity and the time of flight (T):
R = v₀x × T
Since we are looking for the angle θ, we need to find expressions for v₀x and H first.
To find the time of flight, we can use the vertical component of velocity at the highest point in the trajectory (when the projectile reaches its maximum elevation). At this point, the vertical component of velocity is zero. Using this information, we can calculate the time of flight (T) as:
0 = v₀y - g × T/2,
which gives us:
T = 2 × v₀y / g, where g is the acceleration due to gravity.
Now, let's substitute the expressions for v₀x, H, and T into the range equation:
R = (v₀ × cos(θ)) × (2 × v₀y / g)
Rearranging the equation, we get:
H = 2 × v₀y² / g
Since R = 0.5 × H, we can write:
0.5 × H = (v₀ × cos(θ)) × (2 × v₀y / g)
Simplifying the equation, we find:
0.5 × (2 × v₀y² / g) = v₀ × cos(θ) × (2 × v₀y / g)
Canceling out the common terms, we get:
v₀y = v₀ × cos(θ)
Since v₀y = v₀ × sin(θ) (from our earlier definition), we have:
v₀ × sin(θ) = v₀ × cos(θ)
Canceling out the v₀ term, we get:
sin(θ) = cos(θ)
Now, using the trigonometric identity:
sin(θ) = cos(θ) implies θ = 45°
So, the angle between the initial velocity and the ground is 45° above the horizontal.