When a projectile is launched at an initial height of H feet above the ground at an angle of theta with the horizontal and initial velocity is Vo feet per second. the path of the projectile...

Respuesta :

Given,

The initial height of H feet.

The initial velocity of the object is Vo.

The equation of the path of projectile is,

[tex]y=h+x\text{ tan }\theta-\frac{x^2}{2V_0\cos ^2\theta}_{}\text{ }[/tex]

This is the expression of the projectle path.

Hence, the path of the projectile object is y = h + xtan(theta) - x²/2V₀²cos²(theta)