Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

a person in an elevator accelerating upwards with an acceleration of 2m/s2 tosses a coin vertically upwards with a speed of 20 m/s after how much time will the coin fall back into his hand?

3 years ago

Raman Mishra
67 Points

Taking the elevator as the reference frame, we have:
Initial velocity of the coin wrt the elevator = u = 20m/s
Acceleration of the coin wrt the elevator = a = (2-g)m/s2
Since the coin falls back to its original position therefore its displacement = s=0.
Let the time taken be t.
Using the formula  $\dpi{80} s = ut + \frac{1}{2} at^{2}$ ,   we get
$\dpi{80} 0 = 20t + \frac{1}{2} (2-g)t^{2}$
Solving for t, we get  $\dpi{80} \small t= \frac{40}{g-2}$
Taking g=10m/s2 we get t= 5 seconds.
3 years ago
Prachi
29 Points
Taking the elevator as a reference frame. Now since elevator is a non-inertial frame pseudo force is applied. Since elevator is accelerating upwards therefore pseudo force acts in downward direction By drawing the F.B.D of thw coin, summation of force on y-axis give F = ma(psuedoforce) + mg(weight of coin)mA = m(g+a)A = g+aA = 10+2A = 12m/s^2 . Hence new acceleration of coin is 12m/s^2.Now time taken by the coin to come back in the hand is 2t and t =v/ATherefore 2 x 20/12= 40/12t = 3.33 s

3 years ago
nesha leda
15 Points

f=m(g+a)
ma'=m(g+a)
a'=g+a
=10+2=12m/s^2
a'/t=v
t=a'/v
=12/20=3/5m/s^2

hence the answer is  3/5 seconds
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions