a projectile is launched with an initial velocity of 60 m/s at an angle of 30 degrees above the horizontal. how far does it travel?
To find the distance traveled by the projectile, we can break down the initial velocity into its horizontal and vertical components. The horizontal component represents the projectile's velocity in the x-direction, while the vertical component represents its velocity in the y-direction.
Given:
Initial velocity (v₀) = 60 m/s
Angle (θ) = 30°
Step 1: Calculate the horizontal and vertical components of the velocity.
The horizontal velocity (v₀x) is determined using the formula:
v₀x = v₀ * cos(θ)
v₀x = 60 m/s * cos(30°)
The vertical velocity (v₀y) is determined using the formula:
v₀y = v₀ * sin(θ)
v₀y = 60 m/s * sin(30°)
Step 2: Calculate the time taken for the projectile to rise and fall.
Since the projectile follows a symmetrical path, the time taken to reach the peak (t₁) is equivalent to the time taken to fall from the peak (t₂). The total time (t) can be calculated as twice the time taken to reach the peak.
t = 2 * t₁
The time taken to reach the peak (t₁) can be calculated using the formula:
t₁ = v₀y / g
where g represents the acceleration due to gravity, approximately 9.8 m/s².
Step 3: Calculate the maximum height reached by the projectile.
The maximum height (h) can be determined using the formula:
h = v₀y² / (2 * g)
Step 4: Calculate the total flight time (t).
t = 2 * t₁
t = 2 * (v₀y / g)
Step 5: Calculate the horizontal distance traveled (d).
The horizontal distance (d) can be calculated using the formula:
d = v₀x * t
Finally, substitute the values and perform the calculations:
v₀x = 60 m/s * cos(30°)
v₀y = 60 m/s * sin(30°)
t = 2 * (v₀y / g)
h = v₀y² / (2 * g)
d = v₀x * t
By plugging in the values and performing the calculations, you can determine the horizontal distance traveled by the projectile.