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answer asap giving max points and brainliest A projectile is launched at an angle of 30° and travels a distance of 400 meters. Gravitational acceleration equals 9.8 m/sec2 calculate the initial velocity.
answer choices:
67
87
57

Respuesta :

Answer:

67 m/s

Step-by-step explanation:

Formula for range :

  • Range = u²sin2θ / g
  1. u = initial velocity
  2. θ = angle of launch
  3. g = gravitational acceleration

Solving with the given values :

  • 400 = u² x sin60° / 9.8
  • u² x √3/2 = 3920
  • u² = 7840/√3
  • u² = 4526.4
  • u = 67 m/s (approximately)
semsee45

Answer:

[tex]\sf u=67\:ms^{-1}[/tex]

Step-by-step explanation:

Assuming the distance traveled is the horizontal distance.

Horizontal Range Formula

[tex]\sf R=\dfrac{u^2 \sin 2\theta}{g}[/tex]

where:

  • R = horizontal range
  • u = initial velocity
  • [tex]\theta[/tex] = angle of initial velocity
  • g = acceleration due to gravity

Given:

  • R = 400 m
  • [tex]\theta[/tex] = 30°
  • g = 9.8 m/s²

Substituting the given values into the formula and solving for u:

[tex]\implies \sf 400=\dfrac{u^2 \sin 60^{\circ}}{9.8}[/tex]

[tex]\implies \sf 3920=u^2 \sin 60^{\circ}[/tex]

[tex]\implies \sf 3920=\left(\dfrac{\sqrt{3}}{2}\right)u^2[/tex]

[tex]\implies \sf u^2=\dfrac{7840}{\sqrt{3}}[/tex]

[tex]\implies \sf u=\sqrt{\left(\dfrac{7840}{\sqrt{3}} \right) }[/tex]

[tex]\implies \sf u=67.2787196...[/tex]

[tex]\implies \sf u=67\:ms^{-1}\:(nearest\:whole\:number)[/tex]

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