Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

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

An observer sees a flower pot sail up and then back down past a window 2.45 m high. If the total time the pot is in the sight is 1 s, what is the height above the window upto which the pot rises ?

68 Points
10 years ago

Dear  Fawz naim ,

THe time the pot takes to go up and down through the window must be same as the conditions in both cases are same. The magnitude of velocity at the entry and exit points are same and distance travelled is same too . so time taken to go up and down = 0.5 s

now we can take the axis of reference anywhere on the trajectory of the motion , we take it on the bottom of window. Assuming the velociyt at that point to be u , and applying the equaitons of motion ,

2.45 = u*(.5) -(1/2) *9.8*.52

solving we get u = 7.35m/s

so the maximum height from the bottom of the window = u2/2g  = 2.7625

height from the top of window is  2.7625- 2.45 = .306 m

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.