A catapult launches a stone at an angle of 20° above the horizontal at a speed of 26.8 m/s. Assuming the stone begins its trajectory at the origin, determine the distance in meters from the origin that the stone achieves after 1.05 s.
Note: The stone will change position in the x and y direction, so you need to calculate the straight-line distance between the origin and the position of the stone.
Vo = 28.6 m/s[20o]
Xo = 28.6*Cos20 = 25.18 m/s.
Yo = 28.6*sin20 = 9.78 m/s.
Dx = Xo * t = 25.18m/s * 1.05s. = 26.4 m
To determine the distance the stone achieves after 1.05 seconds, we can break down the motion of the stone into its horizontal and vertical components.
First, let's find the horizontal distance covered by the stone after 1.05 seconds.
The horizontal component of the stone's velocity remains constant throughout the motion. We can find this component using the following equation:
Vx = V * cos(θ)
Where Vx is the horizontal component of velocity, V is the initial velocity (26.8 m/s), and θ is the launch angle (20°).
So, the horizontal component of the stone's velocity is:
Vx = 26.8 m/s * cos(20°)
Vx ≈ 25.3 m/s
Next, we can find the horizontal distance covered by the stone in 1.05 seconds using the equation:
x = Vx * t
Where x is the horizontal distance, Vx is the horizontal component of velocity, and t is the time taken (1.05 seconds).
x = 25.3 m/s * 1.05 s
x ≈ 26.6 m
Therefore, the horizontal distance covered by the stone after 1.05 seconds is approximately 26.6 meters.
Now, let's find the vertical distance covered by the stone after 1.05 seconds.
The vertical component of the stone's velocity changes due to gravity. We can find this component using the following equation:
Vy = V * sin(θ)
Where Vy is the vertical component of velocity, V is the initial velocity (26.8 m/s), and θ is the launch angle (20°).
So, the vertical component of the stone's velocity is:
Vy = 26.8 m/s * sin(20°)
Vy ≈ 9.1 m/s
To find the vertical distance covered by the stone in 1.05 seconds, we can use the equation for vertical motion:
y = V0 * t + (1/2) * a * t^2
Where y is the vertical distance, V0 is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).
In this case, the stone starts from rest in the vertical direction, so V0 = 0. Therefore, the equation simplifies to:
y = (1/2) * a * t^2
y = (1/2) * (-9.8 m/s^2) * (1.05 s)^2
y ≈ -5.1 m
The negative sign indicates that the stone is below the initial position.
Finally, we can calculate the straight-line distance between the origin and the position of the stone after 1.05 seconds using the Pythagorean theorem:
Distance = sqrt(x^2 + y^2)
Distance = sqrt((26.6 m)^2 + (-5.1 m)^2)
Distance ≈ 27 meters
Therefore, the stone achieves a distance of approximately 27 meters from the origin after 1.05 seconds.