A projectile is fired into the air from the top of a cliff of height h = 186 m above a valley. Its initial velocity is v0 = 64.8 m/s at an angle θ = 59° above the horizontal. Where does the projectile land? (Ignore any effects due to air resistance.)
See previous post: Thu 12-4-14, 9:49 AM.
To solve this problem, we can break down the initial velocity into its horizontal and vertical components.
1. Horizontal Component: The horizontal component of the initial velocity is given by v₀ₓ = v₀ * cos(θ). Plugging in the values, we get v₀ₓ = 64.8 * cos(59°).
2. Vertical Component: The vertical component of the initial velocity is given by v₀ᵧ = v₀ * sin(θ). Plugging in the values, we get v₀ᵧ = 64.8 * sin(59°).
Now, we can calculate the time of flight of the projectile using the vertical component. The equation to find the time of flight (t) is given by: t = (2 * v₀ᵧ) / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we have t = (2 * 64.8 * sin(59°)) / 9.8.
Next, we can calculate the horizontal distance traveled by the projectile (range). The equation to find the range (R) is given by: R = v₀ₓ * t.
Substituting the values, we have R = (64.8 * cos(59°)) * [(2 * 64.8 * sin(59°)) / 9.8].
By evaluating this expression, we can find the horizontal distance traveled by the projectile.