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An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle of 23 degrees BELOW the horizontal. How long, in seconds, does it take to reach the ground

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Corsaquix

Answer:

[tex]3.58\:\mathrm{s}[/tex]

Explanation:

We can use the kinematics equation [tex]\Delta y=v_it+\frac{1}{2}at^2[/tex] to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

[tex]\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}[/tex]

Now we can substitute values in our kinematics equation:

  • [tex]\Delta y=-100[/tex]
  • [tex]a=-9.8\:\mathrm{m/s^2}[/tex] (acceleration due to gravity)
  • [tex]v_i=-10.3283743813\:\mathrm{m/s}[/tex]
  • Solving for [tex]t[/tex]

[tex]-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}[/tex]

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