A stone is thrown from ground level at 84 m/s. It's speed when it reaches it's highest point is 50 m/s. Find the angle, above the horizontal, of the stones initial velocity.
v(x) =v₀(x) = 50 m/s
v₀=84 m/s
cosα =v₀(x)/ v₀=50/84 =0.6
α=53.5⁰
how did you get 53.5
To find the angle above the horizontal of the stone's initial velocity, we can use trigonometry and consider the motion of the stone as a projectile.
Let's break down the motion of the stone into vertical and horizontal components. The initial velocity can be decomposed into two parts: one in the horizontal direction and one in the vertical direction.
Given:
Initial speed (magnitude of initial velocity), v₀ = 84 m/s
Speed at the highest point, v = 50 m/s
1. Determine the initial vertical and horizontal components of the velocity using the given information.
The horizontal component of velocity (vₓ) remains constant throughout the stone's flight because there are no horizontal forces acting on it. Therefore, the horizontal component can be expressed as:
vₓ = v₀ * cosθ
where θ is the angle above the horizontal.
The vertical component of velocity (vᵧ) changes due to the acceleration due to gravity. At the highest point, the vertical velocity is zero. Therefore, the vertical component can be expressed as:
vᵧ = 0
2. Using the components of the velocity, determine the angle θ.
To find the unknown angle θ, we can rearrange the equation for the horizontal component of velocity:
vₓ = v₀ * cosθ
Divide both sides by v₀:
cosθ = vₓ / v₀
Now, substitute the known values:
cosθ = 50 m/s / 84 m/s
Calculate the value of cosθ:
cosθ ≈ 0.5952
Take the inverse cosine (cos⁻¹) to find θ:
θ ≈ cos⁻¹(0.5952)
θ ≈ 53.1°
Therefore, the angle above the horizontal of the stone's initial velocity is approximately 53.1°.