A soccer ball is kicked at 72.0 km/h at 20o above horizon. How far from initial spot it was kicked, is the ball 2.00 m above the ground? (the question was given to us like this, unsure of what it means asking if the ball is 2m above ground)
The Correct Answer and Explanation is:
To solve this problem, we need to break the motion of the soccer ball into horizontal and vertical components and apply the kinematic equations of projectile motion.
Step 1: Convert Units and Set Up Variables
First, convert the speed from km/h to m/s:72.0 km/h=72.0×10003600 m/s=20.0 m/s72.0 \, \text{km/h} = \frac{72.0 \times 1000}{3600} \, \text{m/s} = 20.0 \, \text{m/s}72.0km/h=360072.0×1000m/s=20.0m/s
We are given the following:
- Initial speed (v0v_0v0) = 20.0 m/s
- Angle of launch (θ\thetaθ) = 20°
- Gravitational acceleration (ggg) = 9.81 m/s²
- Vertical displacement (yyy) = 2.00 m (we need to check if the ball is 2 m above the ground)
Step 2: Resolve Initial Velocity into Components
The initial velocity components are:
- Horizontal velocity:
v0x=v0cos(θ)=20.0×cos(20∘)=20.0×0.9397=18.794 m/sv_{0x} = v_0 \cos(\theta) = 20.0 \times \cos(20^\circ) = 20.0 \times 0.9397 = 18.794 \, \text{m/s}v0x=v0cos(θ)=20.0×cos(20∘)=20.0×0.9397=18.794m/s
- Vertical velocity:
v0y=v0sin(θ)=20.0×sin(20∘)=20.0×0.3420=6.840 m/sv_{0y} = v_0 \sin(\theta) = 20.0 \times \sin(20^\circ) = 20.0 \times 0.3420 = 6.840 \, \text{m/s}v0y=v0sin(θ)=20.0×sin(20∘)=20.0×0.3420=6.840m/s
Step 3: Find Time When the Ball is 2 m Above the Ground
The equation for vertical displacement is:y=v0yt−12gt2y = v_{0y} t – \frac{1}{2} g t^2y=v0yt−21gt2
Substituting y=2.00y = 2.00y=2.00 m, v0y=6.840v_{0y} = 6.840v0y=6.840 m/s, and g=9.81g = 9.81g=9.81 m/s²:2.00=6.840t−12(9.81)t22.00 = 6.840t – \frac{1}{2}(9.81)t^22.00=6.840t−21(9.81)t2
This is a quadratic equation:4.905t2−6.840t+2.00=04.905t^2 – 6.840t + 2.00 = 04.905t2−6.840t+2.00=0
Using the quadratic formula:t=−(−6.840)±(−6.840)2−4(4.905)(2.00)2(4.905)t = \frac{-(-6.840) \pm \sqrt{(-6.840)^2 – 4(4.905)(2.00)}}{2(4.905)}t=2(4.905)−(−6.840)±(−6.840)2−4(4.905)(2.00)t=6.840±46.786−39.249.81t = \frac{6.840 \pm \sqrt{46.786 – 39.24}}{9.81}t=9.816.840±46.786−39.24t=6.840±7.5469.81t = \frac{6.840 \pm \sqrt{7.546}}{9.81}t=9.816.840±7.546t=6.840±2.759.81t = \frac{6.840 \pm 2.75}{9.81}t=9.816.840±2.75
So, two possible values for time:t1=6.840+2.759.81=9.5909.81=0.978 st_1 = \frac{6.840 + 2.75}{9.81} = \frac{9.590}{9.81} = 0.978 \, \text{s}t1=9.816.840+2.75=9.819.590=0.978st2=6.840−2.759.81=4.0909.81=0.417 st_2 = \frac{6.840 – 2.75}{9.81} = \frac{4.090}{9.81} = 0.417 \, \text{s}t2=9.816.840−2.75=9.814.090=0.417s
We take the larger time, t1=0.978t_1 = 0.978t1=0.978 s, because this is when the ball reaches 2 m for the second time during its flight.
Step 4: Find Horizontal Distance
Now, we can find the horizontal distance by using the formula:x=v0x×tx = v_{0x} \times tx=v0x×tx=18.794 m/s×0.978 s=18.41 mx = 18.794 \, \text{m/s} \times 0.978 \, \text{s} = 18.41 \, \text{m}x=18.794m/s×0.978s=18.41m
Conclusion
The soccer ball will be 2.00 m above the ground at a horizontal distance of 18.41 m from the initial spot where it was kicked.
This confirms the horizontal distance when the ball is at 2 m height. The question likely wants to know where along its trajectory the ball is exactly 2 m above the ground.
