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Projectile trajectory y = ax - bx^2: angle, maximum height or initial velocity

This type of question, asked 3 times: 2021 Paper 26, 26 Jul 2022, 8 Apr 2023. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 26

    2
    The trajectory a projectile in a vertical plane is y=αx−βx2y = \alpha x - \beta x^{2}, where α\alpha and β\beta are constants and xx &y\& y are respectively the horizontal and vertical distance of the projectile from the point of projection. The angle of projection θ\theta and the maximum height attained HH are respectively given by:

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  2. Question 2

    26 Jul 2022

    21
    If the initial velocity in horizontal direction of a projectile is unit vector i^\widehat{i} and the equation of trajectory is y=5x(1−x)y = 5x(1 - x). The yy component of the initial velocity is ____\_\_\_\_ m/sm/s. (Take g=10 m/s2g = 10\text{ }m/s^{2} )

  3. Question 3

    8 Apr 2023

    95
    The trajectory of projectile, projected from the ground is given by y=x−x220y = x -\frac{x^{2}}{20}. Where xx and yy are measured in meter. The maximum height attained by the projectile will be.

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