For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substantiate your answer with mathematical support.
20.5 m/s
To find the angle at which the maximum horizontal distance is possible in projectile motion, we can make use of the formula for the horizontal range or distance:
Range = (v^2 * sin(2θ))/g
Where:
- Range is the horizontal distance traveled by the projectile
- v is the velocity of projection
- θ is the angle of projection
- g is the acceleration due to gravity
Now, we can find the angle θ at which the maximum horizontal distance occurs by differentiating the range formula with respect to θ and setting it equal to zero:
d(Range)/dθ = (v^2 * cos(2θ))/g = 0
Since the numerator cannot be equal to zero, we can ignore it and solve for cos(2θ) = 0:
cos(2θ) = 0
The solutions to this equation are θ = 45° and θ = 135°. However, projectile motion has a symmetrical nature, meaning that the path traced by the projectile is the same for θ and (180° - θ).
Therefore, for any given velocity of projection, the maximum horizontal distance is possible when θ = 45° because both 45° and (180° - 45° = 135°) produce symmetrical paths.