Respuesta :
Given:
v = 11 m/s, the launch speed
h = 3.6 m, the height of the ceiling
Assume g = 9.8 m/s², and neglect air resistance.
Let θ = the launch angle, measured above the horizontal.
The initial vertical speed is
u = 11 sin θ m/s
At the maximum height of 3.6 m, the vertical velocity is zero.
Therefore
(11 sinθ m/s)² - 2*(9.8 m/s²)*(3.6 m) = 0
121 sin² θ = 70.56
sin²θ = 0.5831
sin θ = 0.7636
θ = sin⁻¹ 0.7636 = 49.8° ≈ 50°
Answer: 50° above the horizontal
v = 11 m/s, the launch speed
h = 3.6 m, the height of the ceiling
Assume g = 9.8 m/s², and neglect air resistance.
Let θ = the launch angle, measured above the horizontal.
The initial vertical speed is
u = 11 sin θ m/s
At the maximum height of 3.6 m, the vertical velocity is zero.
Therefore
(11 sinθ m/s)² - 2*(9.8 m/s²)*(3.6 m) = 0
121 sin² θ = 70.56
sin²θ = 0.5831
sin θ = 0.7636
θ = sin⁻¹ 0.7636 = 49.8° ≈ 50°
Answer: 50° above the horizontal
[tex]50^\circ[/tex] angle above the horizontal.
Step by Step Solution :
Given :
Launch speed = 11 m/sec
Height of the ceiling = 3.6 m
[tex]\rm g = 9.8\;m/sec^2[/tex]
Calculation :
Let [tex]\theta[/tex] be the launch angle, measured above the horizontal.
Vertical component of initial speed is
[tex]\rm v = 11sin\theta[/tex]
We know that
[tex]\rm v^2 - u^2=2as[/tex]
[tex]\rm 11^2 sin^2\theta - 2\times9.8\times 3.6 =0[/tex]
[tex]\rm sin\theta = 0.7636[/tex]
[tex]\rm \theta = sin^-^10.7636[/tex]
[tex]\theta = 49.8^\circ \approx 50^\circ[/tex]
[tex]50^\circ[/tex] angle above the horizontal they launched to just graze the ceiling.
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