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

Answer:

76°

Explanation:

In projectile,

Maximum height = $\frac{u^{2}sin^{2} \theta }{2g}$ and range = $\frac{u^{2}sin2\theta }{g}$

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;

$\frac{u^{2}sin^{2} \theta }{2g} = \frac{u^{2}sin2\theta }{g}$

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

$\frac{u^{2}sin^{2} \theta }{2g} = \frac{u^{2}2sin\theta cos\theta }{g}\\\frac{sin\theta}{2} =2cos\theta \\sin\theta = 4cos\theta\\\frac{sin\theta}{cos\theta} = 4\\tan\theta = 4\\\theta = tan^{-1} \\4\theta = 75.9^{0}$

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