Respuesta :
Answer:
To find the maximum height the shuttlecock will reach, we can use the following kinematic equation:
\[ \text{Final velocity}^2 = \text{Initial velocity}^2 - 2 \times \text{acceleration} \times \text{vertical distance} \]
Since the shuttlecock will reach its maximum height, its final vertical velocity will be 0. The acceleration due to gravity is approximately \(9.81 \, \text{m/s}^2\). The initial vertical velocity (\(v_{i_y}\)) can be found using trigonometry:
\[ v_{i_y} = v_i \times \sin(\theta) \]
Where:
\(v_i\) = initial speed = 10 m/s
\(\theta\) = launch angle = 35°
Plugging in the values:
\[ v_{i_y} = 10 \, \text{m/s} \times \sin(35°) \approx 10 \, \text{m/s} \times 0.5736 \approx 5.736 \, \text{m/s} \]
Now, we can use the kinematic equation to find the maximum height (\(h\)):
\[ 0 = (5.736 \, \text{m/s})^2 - 2 \times (-9.81 \, \text{m/s}^2) \times h \]
\[ 0 = 32.89 \, \text{m}^2/\text{s}^2 + 19.62 \, \text{m/s}^2 \times h \]
\[ -32.89 \, \text{m}^2/\text{s}^2 = 19.62 \, \text{m/s}^2 \times h \]
\[ h = \frac{-32.89 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \approx -1.677 \, \text{m} \]
Since the height cannot be negative, we made a mistake somewhere. Let's correct it:
\[ h = \frac{(-32.89 \, \text{m}^2/\text{s}^2)}{(-19.62 \, \text{m/s}^2)} \approx 1.678 \, \text{m} \]
So, the shuttlecock will reach a maximum height of approximately 1.678 meters.
To find the horizontal distance traveled before being contacted by the opponent's racket, we can use the time of flight. The time of flight can be found using the equation:
\[ \text{Time of flight} = \frac{2 \times v_{i_y}}{g} \]
\[ \text{Time of flight} = \frac{2 \times 5.736 \, \text{m/s}}{9.81 \, \text{m/s}^2} \approx \frac{11.472 \, \text{s}}{9.81} \approx 1.169 \, \text{s} \]
Now, we can find the horizontal distance (\(d_x\)) traveled using the equation:
\[ d_x = v_{i_x} \times \text{Time of flight} \]
Where \(v_{i_x}\) is the initial horizontal velocity:
\[ v_{i_x} = v_i \times \cos(\theta) \]
\[ v_{i_x} = 10 \, \text{m/s} \times \cos(35°) \approx 10 \, \text{m/s} \times 0.8192 \approx 8.192 \, \text{m/s} \]
\[ d_x = 8.192 \, \text{m/s} \times 1.169 \, \text{s} \approx 9.582 \, \text{m} \]
Therefore, the shuttlecock will travel approximately 9.582 meters horizontally before being contacted by the opponent's racket.
