Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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

Sagot :

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]

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.