A ball is shot lip an incline plane with an intial velocity of \( 40 \mathrm{~m} / \mathrm{s} \), the incline plane has iriclination of \( 25^{\circ} \). Calculate (a) maximum distance travelled (a) Time taken to do this souiney c) Time of Light \[ g=10 \mathrm{~m} / \mathrm{s} \]

To solve the problem, we'll break it down into parts based on the equations of motion for projectile motion. ### Given Data: - **Initial Velocity, \( v_0 \) = 40 m/s** - **Inclination, \( \theta = 25^\circ \)** - **Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)** ### (a) Maximum Distance Travelled: First, we need to find the horizontal and vertical components of the initial velocity: - **Horizontal Component:** \[ v_{0x} = v_0 \cdot \cos(\theta) = 40 \cdot \cos(25^\circ) \] \[ v_{0x} \approx 40 \cdot 0.9063 \approx 36.25 \, \text{m/s} \] - **Vertical Component:** \[ v_{0y} = v_0 \cdot \sin(\theta) = 40 \cdot \sin(25^\circ) \] \[ v_{0y} \approx 40 \cdot 0.4226 \approx 16.90 \, \text{m/s} \] Now, to find the time taken to reach the maximum height (where vertical velocity becomes zero): \[ t_{up} = \frac{v_{0y}}{g} = \frac{16.90}{10} \approx 1.69 \, \text{s} \] The total time of flight is twice this value (up and down): \[ T = 2 \cdot t_{up} \approx 2 \cdot 1.69 \approx 3.38 \, \text{s} \] Now, we can find the maximum distance travelled along the incline. The horizontal distance covered can be calculated as: \[ \text{Distance} = v_{0x} \cdot T \approx 36.25 \cdot 3.38 \approx 122.00 \, \text{m} \] ### Summary of Answers: (a) The maximum distance travelled along the incline is approximately \( 122.00 \, \text{m} \). (b) The time taken for this journey is approximately \( 3.38 \, \text{s} \). (c) The total time of flight is also \( 3.38 \, \text{s} \).