Respuesta :
Answer:
a) 76º b) Rmax = 2.1R
Explanation:
When reading the exercise you can see that it is a projectile launching exercise, let's write the equations for reach and height.
R = Vo2 sin 2T / g
VY² = Voy² -2gY
The condition they give us
R = Ym
At the point of maximum height the vertical speed is zero (Vy = 0)
0 = Voy2 -2 gY
Y = I'm going 2 / 2g
We match the equations
Voy² /2g = Vo² sin 2θ / g
Vo = Vo sin T
(Vo² sin²θ)/2 = vo² sin 2θ
sin² θ = 2 sin 2θ
Use the relationship. sin 2θ = 2 sin θ cos θ
sin² θ = 2 (2 sin θ cos θ)
sin θ = 4 cos θ
Tan θ = 4
θ = tan⁻¹ 4
θ = 76º
b) to find the maximum range we analyze the expression of the range and see that it will be maximum when the sin 2θ is 1 which implies an angle of 45º, in this case the equation is
Rmax = Vo² /g maximum range
R = Vo²/g sin 2θ
To find the relationship
R / Rmax = sin 2T
Rmax = R / sin (2 76)
Rmax = 2.1 R
c) When reviewing expressions we can see that when equalizing the scope and maximum height the value of the acceleration of gravity is canceled, because the answer does not depend on the severity, consequently, if the experiment is performed in another plan with different acceleration of severity the result is not altered.
