a projectile has an initial velocity of 15.0 m/s at an angle of 30 degrees above the horizontal. What is the location of the projectile 2.0 seconds later?
The time of flight of the projectle is
2 Vo*sin30/g = 1.53 seconds
After 2.0 seconds, it will have hit the ground already. Whether and how far it rolled after that I cannot say.
X = (V COSα)t
= 15*cos30*2
= 25.98
To find the location of the projectile 2.0 seconds later, we need to break down the initial velocity into its horizontal and vertical components.
Step 1: Find the horizontal component of the initial velocity:
Horizontal velocity (Vx) = Initial velocity (Vi) * cos(angle)
Vx = 15.0 m/s * cos(30°)
Vx = 15.0 m/s * 0.866
Vx = 12.99 m/s (rounded to two decimal places)
Step 2: Find the vertical component of the initial velocity:
Vertical velocity (Vy) = Initial velocity (Vi) * sin(angle)
Vy = 15.0 m/s * sin(30°)
Vy = 15.0 m/s * 0.5
Vy = 7.5 m/s
Step 3: Find the horizontal displacement after 2.0 seconds:
Horizontal displacement (Dx) = Vx * time
Dx = 12.99 m/s * 2.0 s
Dx = 25.98 m (rounded to two decimal places)
Step 4: Find the vertical displacement after 2.0 seconds:
Vertical displacement (Dy) = (Vy * time) + (0.5 * acceleration * time^2)
Given that the acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of the positive vertical direction):
Dy = (7.5 m/s * 2.0 s) + (0.5 * -9.8 m/s^2 * (2.0 s)^2)
Dy = 15.0 m + (-19.6 m)
Dy = -4.6 m (rounded to one decimal place)
Step 5: Find the final location of the projectile:
The final location will consist of the horizontal and vertical displacements.
Final location = (Dx, Dy)
Final location = (25.98 m, -4.6 m)
Therefore, the location of the projectile 2.0 seconds later is approximately (25.98 m, -4.6 m).
To find the location of the projectile 2.0 seconds later, we need to break down the initial velocity into its horizontal and vertical components.
The horizontal component of velocity (Vx) can be calculated using the formula:
Vx = V * cos(θ)
where:
V = initial velocity = 15.0 m/s
θ = angle above the horizontal = 30 degrees
Substituting the given values into the formula, we get:
Vx = 15.0 m/s * cos(30 degrees)
Vx = 15.0 m/s * (√3/2)
Vx ≈ 12.99 m/s
The vertical component of velocity (Vy) can be calculated using the formula:
Vy = V * sin(θ)
where:
V = initial velocity = 15.0 m/s
θ = angle above the horizontal = 30 degrees
Substituting the given values into the formula, we get:
Vy = 15.0 m/s * sin(30 degrees)
Vy = 15.0 m/s * (1/2)
Vy = 7.50 m/s
Now that we have the horizontal and vertical components of velocity, we can calculate the horizontal and vertical displacements at a given time using the following equations:
Horizontal displacement:
dx = Vx * t
where:
Vx = horizontal component of velocity = 12.99 m/s (from earlier)
t = time = 2.0 seconds
Substituting the given values into the formula, we get:
dx = 12.99 m/s * 2.0 s
dx ≈ 25.98 m
Vertical displacement:
dy = Vy * t + (1/2) * g * t^2
where:
Vy = vertical component of velocity = 7.50 m/s (from earlier)
t = time = 2.0 seconds
g = acceleration due to gravity = 9.81 m/s^2
Substituting the given values into the formula, we get:
dy = 7.50 m/s * 2.0 s + (1/2) * 9.81 m/s^2 * (2.0 s)^2
dy ≈ 15.0 m + 19.62 m
dy ≈ 34.62 m
Now, we can find the location of the projectile by combining the horizontal and vertical displacements. Assuming the initial position is at the origin (0,0), the location of the projectile 2.0 seconds later would be (25.98 m, 34.62 m).