Question

Para que ángulo el alcance es igual a la altura máxima en un lanzamiento de proyectil ideal sobre un plano horizontal. Translation: For which angle the range equals the maximum height in an ideal projectile launch on a horizontal plane. a) 45º b) 60º c) 76º d) 90º

Answer 1

76°

Step-by-step explanation:

In projectile,

Maximum height =
`(u^(2)sin^(2) \theta )/(2g)` and range =
`(u^(2)sin2\theta )/(g)`

u = velocity of the body

`\theta` = angle of launch

g = acceleration due to gravity

If the range equals the maximum height in an ideal projectile launch on a horizontal plane, this means;

`(u^(2)sin^(2) \theta )/(2g) = (u^(2)sin2\theta )/(g)`

since
`sin2\theta = 2sin\theta cos\theta`

`(u^(2)sin^(2) \theta )/(2g) = (u^(2)2sin\theta cos\theta )/(g)\\(sin\theta)/(2) =2cos\theta \\sin\theta = 4cos\theta\\(sin\theta)/(cos\theta) = 4\\tan\theta = 4\\\theta = tan^(-1) \\4\theta = 75.9^(0)`

The angle at which the range equals the maximum height in an ideal projectile launch on a horizontal plane is approximately 76°