At what angle is an object launched if its initial vertical speed is 3.75 m/s and its initial horizontal speed is 4.50 m/s?
tanA = 3.75/4.5 = 0.8333,
A = 39.8 deg.
To find the launch angle, we can use the trigonometric relationships between the horizontal and vertical components of the velocity.
Let's assume the launch angle as θ.
The vertical component of the velocity is given by:
v_y = v_initial * sin(θ)
The horizontal component of the velocity is given by:
v_x = v_initial * cos(θ)
Given that the initial vertical speed is 3.75 m/s and the initial horizontal speed is 4.50 m/s, we can set up the following equations:
3.75 = v_initial * sin(θ) (Equation 1)
4.50 = v_initial * cos(θ) (Equation 2)
To solve for θ, we can divide Equation 1 by Equation 2:
3.75 / 4.50 = (v_initial * sin(θ)) / (v_initial * cos(θ))
0.833 = tan(θ)
Now, we can take the inverse tangent (or arctan) of both sides to find the launch angle:
θ = arctan(0.833)
Using a calculator, we find:
θ ≈ 39.8 degrees
Therefore, an object is launched at an angle of approximately 39.8 degrees.
To determine the angle at which the object is launched, we can use trigonometry. Given the initial vertical speed (Vy) and initial horizontal speed (Vx), we can use the formula:
tan(θ) = Vy / Vx
where θ is the launch angle.
First, plug in the given values:
tan(θ) = 3.75 m/s / 4.50 m/s
Now, we can calculate the angle by taking the inverse tangent (arctan) of both sides of the equation:
θ = arctan(3.75 m/s / 4.50 m/s)
Using a scientific calculator, compute the arctan:
θ ≈ 38.66 degrees
Therefore, the object is launched at an angle of approximately 38.66 degrees.