Respuesta :
Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
Answer:
The correct option is;(C) greater than zero but less than 45° above the horizontal
Explanation:
Here, we have for maximum horizontal distance
h = v×t
Where t is the time of flight
The time of flight is given by
s = v·t - 0.5 × gt² to maximize the time of flight, we therefore increase the height such that
Since the range is given by
Horizontal range, x = v·t·cosα
Vertical range, y = v·t·sinα - 0.5·g·t²
When the particle comes back to initial level, we have
0 = v·t·sinα - 0.5·g·t² → 0 = t(v·sinα - 0.5·g·t)
So that t = 0 or t = [tex]\frac{2\cdot v\cdot sin\alpha }{g}[/tex]
Therefore, horizontal range =
[tex]\frac{v^{2} \cdot 2\cdot cos\alpha \cdot sin\alpha }{g} = \frac{v^2sin(2\alpha) }{g}[/tex]
Therefore maximum range is obtained when α = 45° as sin 90° = 1