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PLS i really need help.
•you decided to play Flick the coin with your friend. The goal of the game is to flick the coin from the top of a table and let it land of the floor as far as possible. The coin that you flicked left the table (0.8m high) with an initial velocity of 5m/s. Compute the following.
A. Time It will take for the coin to reach the ground.
B. Range (how far from the base of the table will the coin land.with solution​

Respuesta :

anvitanvar032

Answer:

A) The time taken by the coin to reach the ground will be 0.4 seconds

B) Range of the motion of the coin will be 2m

Explanation:

This is a case of horizontal projectile motion as there's no initial vertical velocity given to the coin and has only horizontal velocity initially.

According to the equations of Horizontal projectile

  • [tex]Sy=\frac{1}{2} *g*t^2[/tex] {[tex]Sy =[/tex] Vertical distance traveled , [tex]g=[/tex] acceleration due to gravity of Earth  , [tex]t=[/tex] time of flight i.e time taken)
  • [tex]Sx= Vx*t[/tex] ([tex]Sx=[/tex] Range of the projectile ,[tex]Vx=[/tex] initial horizontal velocity, [tex]t=[/tex] time of flight)

Given :

Vertical distance to be traveled  ([tex]Sy[/tex])= 0.8m

Initial horizontal velocity ([tex]Vx[/tex])= 5m/s

For calculations, we will take acceleration due to gravity ([tex]g[/tex]) to be [tex]10 m/s^2[/tex]

A) Putting values in equation [tex]Sy=\frac{1}{2} *g*t^2[/tex]

   ⇒ [tex]0.8=\frac{1}{2} *10*t^2[/tex]

   ⇒[tex]\frac{1.6}{10} =t^2[/tex]

   ⇒ [tex]\sqrt{0.16} =t[/tex]

    ⇒  [tex]0.4=t[/tex]

Therefore, time taken by the coin to reach the ground is 0.4 seconds

B) Now using t= 0.4s putting values in equation [tex]Sx= Vx*t[/tex]

⇒[tex]Sx=5*0.4[/tex]

⇒[tex]Sx=2[/tex]

Hence , Range of the coin is 2m

Learn more about Projectile motion here :

https://brainly.com/question/24216590

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