Respuesta :
Answer:
θ = 14.54°/2 = 7.27°
Therefore, the projection angle θ = 7.27°
Step-by-step explanation:
The range of a projectile can be written as;
R = (v^2 . Sin2θ)/g .....1
Where;
R = range = 37m
v = initial speed = 38m/s
θ = projection angle
g = acceleration due to gravity = 9.8m/s^2
Making sin2θ the subject of formula in equation 1
Sin2θ = Rg/v^2
Sin2θ = (37×9.8)/38^2
Sin2θ = 0.2511
2θ = sininverse(0.2511)
2θ = 14.54°
θ = 14.54°/2 = 7.27°
Therefore, the projection angle θ = 7.27°
The projectile angle of the arrow to hit the target at 37 m away must be 14.558°.
Given to us
Range, R = 37 m
Initial velocity, u = 38 m/s
What is Range?
In projectile motion, the range is the horizontal distance traveled by the object.
[tex]\rm R =\dfrac{u^2\cdot sin\theta}{g}[/tex]
substitute the value,
[tex]37 =\dfrac{(38)^2\cdot sin\theta}{9.81}[/tex]
[tex]sin\theta= 0.25136\\\\\theta = sin^{-1} 0.25136\\\\\theta = 14.558^o[/tex]
Hence, the projectile angle of the arrow to hit the target at 37 m away must be 14.558°.
Learn more about Projectile Motion:
https://brainly.com/question/11049671
