The trajectory of a projectile in a vertical plane is y = α x - β x 2 , where α and β are constants and x   &   y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum…

Question

The trajectory of a projectile in a vertical plane is y = &#945; x - &#946; x 2 , where &#945; and &#946; are constants and x &#160; &#38; &#160; y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection &#952; and the maximum height attained H are respectively given by

Aicharya · Accepted Answer

y = &#945; x - &#946; x 2 comparing with trajectory equation y = x tan &#952; - 1 2 g x 2 u 2 cos 2 &#952; tan &#952; = &#945; &#8658; &#952; = tan - 1 &#945; &#946; = 1 2 g u 2 cos 2 &#952; u 2 = g 2 &#946; cos 2 &#952; Maximum height : H H = u 2 sin 2 &#952; 2 &#160; g = g 2 &#946; cos 2 &#952; sin 2 &#952; 2 &#160; g H = tan 2 &#952; 4 &#946; = &#945; 2 4 &#946;

The trajectory of a projectile in a vertical plane is y = α x - β x 2 , where α and β…

Solution

The trajectory of a projectile in a vertical plane is y = &#945; x - &#946; x 2 , where &#945; and &#946;…

Solution

The trajectory of a projectile in a vertical plane is y = α x - β x 2 , where α and β…