A projectile is fired such that its horizontal range is equal to 2.5 times its maximum height. Find the angle the projectile was fired at
http://answers.yahoo.com/question/index?qid=20091204052014AAr0lCr
it will be 28.25 degree
To find the angle at which the projectile was fired, we can start by breaking down the motion of the projectile into its horizontal and vertical components.
Let's denote the initial velocity of the projectile as v0 and the angle at which it was fired as θ.
1. Horizontal Component:
The horizontal velocity (Vx) of the projectile remains constant throughout its motion, as there are no horizontal forces acting on it. Therefore, we can express the horizontal distance traveled by the projectile as:
Range = Vx * Time
Since the projectile covers twice the distance in the horizontal direction compared to the vertical direction (2.5 times the maximum height), we can write:
Range = 2.5 * Max Height
Now, substitute the values for the horizontal velocity and the maximum height with their respective equations:
Range = (v0 * cos(θ)) * Time
2.5 * Max Height = (v0 * sin(θ)) * Time
2. Vertical Component:
The vertical motion of the projectile can be described by the equations of motion. Considering the projectile is launched from the ground, the initial vertical velocity (Vy) is given by:
Vy = v0 * sin(θ)
Using this initial vertical velocity, we can determine the time of flight (T) and the maximum height (H) reached by the projectile:
T = 2 * Vy / g
H = Vy^2 / (2 * g)
where g is the acceleration due to gravity.
3. Equating Range and Height equations:
By equating the range equation from step 1 with the height equation from step 2, we can eliminate the time variable:
2.5 * H = (v0 * cos(θ)) * (2 * Vy / g)
4. Solving for θ:
Now, substitute the expressions for Vy and H:
2.5 * (v0 * sin(θ))^2 / (2 * g) = (v0 * cos(θ)) * (2 * (v0 * sin(θ)) / g)
Simplify the equation by canceling v0 and dividing both sides by 2:
1.25 * sin(θ)^2 = cos(θ) * sin(θ)
Divide both sides by sin(θ):
1.25 * sin(θ) = cos(θ)
Rearrange the equation:
sin(θ) / cos(θ) = 1.25
Using the identity tan(θ) = sin(θ) / cos(θ):
tan(θ) = 1.25
Now, take the inverse tangent (arctan) of both sides to find the angle θ:
θ = arctan(1.25)
Calculating the inverse tangent of 1.25 using a calculator gives us:
θ ≈ 51.34 degrees
Therefore, the angle at which the projectile was fired is approximately 51.34 degrees.