A soccer ball was kicked off a cliff that was h = 8.4 m above the ground. The ball was kicked at a speed of 25 m s-1 and an angle of 20 to the horizon.
What is magnitude of the horizontal displacement of the ball? Use g = 9.8 m s-2.
See previous post: Thu, 12-15-16, 9:33 PM.
To find the magnitude of the horizontal displacement of the ball, we need to break down the initial velocity into its horizontal and vertical components.
Given:
- Initial speed of the ball (v) = 25 m/s
- Angle (θ) = 20°
- Acceleration due to gravity (g) = 9.8 m/s²
Step 1: Find the horizontal component of the initial velocity (v₀x).
v₀x = v * cos(θ)
v₀x = 25 m/s * cos(20°)
v₀x = 25 m/s * 0.9397
v₀x ≈ 23.4925 m/s
Step 2: Find the time it takes for the ball to reach the ground (t).
Use the vertical component of the initial velocity (v₀y) and the acceleration due to gravity (g).
v₀y = v * sin(θ)
v₀y = 25 m/s * sin(20°)
v₀y = 25 m/s * 0.3420
v₀y ≈ 8.55 m/s
Using the kinematic equation: h = v₀yt - 1/2 * g * t²
Rearranging the equation to solve for t:
t = sqrt((2 * h) / g)
t = sqrt((2 * 8.4 m) / 9.8 m/s²)
t ≈ sqrt(1.7143 s²)
t ≈ 1.31 s (approximate to two decimal places)
Step 3: Find the horizontal displacement (x).
x = v₀x * t
x = 23.4925 m/s * 1.31 s
x ≈ 30.74 m
Therefore, the magnitude of the horizontal displacement of the ball is approximately 30.74 meters.