Physics, asked by mahfadhu395, 5 months ago

A person in an elevator accelerating upwards with an acceleration of 2m/s2, tosses a coin vertically upwards with a speed of 20m/s. After how much time will the coin fall back into his hand? (Take g= 10m/s2)

Answers

Answered by Atαrαh
7

Solution  :

As per the given data ,

  • Acceleration of the elevator = 2 m/s ²
  • Acceleration due to gravity = 10 m/s²

Net acceleration of the coin (a)= 10 + 2 = -  12 m/s ( as the coin is moving against gravity )

  • Initial velocity of the coin = 20 m/s
  • Final velocity ( at the highest point ) = 0 m/s

As the coin is moving with uniform acceleration throughout it's motion we can use the first equation of motion in order to find time .

  • v = u + at

Now let's substitute the given values ,

⇒ 0 = 20 - 12 x t

⇒ t = 20 / 12

⇒ t = 1.6 s (approx.)

We know that ,

  • Time of ascent = Time of descent = 1.6s

Hence ,

The total time taken by the ball to fall back into hand of the person

= 1.6 + 1.6

= 3.2 s

The total time taken by the ball to fall back into hand of the person is 3.2 s .

Similar questions