A ball is kicked at a 45* angle over a 6 ft fence 20 m away at 17 m/s. How far above the fence will the ball pass? No air resistance
To find the height above the fence where the ball will pass, we can use the kinematic equations of motion.
The given information for this problem is:
- Initial velocity (vi) = 17 m/s
- Launch angle (θ) = 45 degrees
- Distance to the fence (x) = 20 m
Step 1: Decompose the initial velocity into horizontal and vertical components.
The horizontal component (vix) will remain constant throughout the ball's flight as there is no air resistance. Thus, vix = vi * cos(θ).
The vertical component (viy) can be calculated using viy = vi * sin(θ).
Given that θ = 45 degrees and vi = 17 m/s:
vix = 17 m/s * cos(45) = 17 m/s * √2 / 2 = 12.02 m/s (approx.)
viy = 17 m/s * sin(45) = 17 m/s * √2 / 2 = 12.02 m/s (approx.)
Step 2: Calculate the time of flight.
The time it takes for the ball to reach the fence can be obtained using the equation x = vix * t. Rearranging the equation, we get t = x / vix.
Given that x = 20 m and vix = 12.02 m/s:
t = 20 m / 12.02 m/s = 1.66 seconds (approx.)
Step 3: Determine the vertical distance above the fence.
To find the height above the fence where the ball will pass, we need to find the maximum height reached by the ball.
Using kinematic equations, we can use the following equation to find the maximum vertical height (hmax) reached by the ball:
hmax = (viy^2) / (2 * g)
Given that viy = 12.02 m/s and g (acceleration due to gravity) = 9.8 m/s^2:
hmax = (12.02 m/s)^2 / (2 * 9.8 m/s^2) ≈ 7.37 m
Therefore, the ball will pass approximately 7.37 meters above the fence.