#### 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

# A falling stone takes 0.2 sec to fall past a window which is 1 m high.from how far above the top of the window was the stone dropped

Vikas TU
14149 Points
4 years ago
Let V0 = the underlying speed of the stone as it crossed the top edge of the window
Let h = the tallness over the window that the stone was dropped with no underlying speed.
Let g = the speeding up because of gravity = 9.8 m/s²
Let t = the time that it takes to fall past the window = 0.2 s
Let d = the tallness of the window = 1 m
Utilizing the separation condition for steady speeding up with an underlying speed:
d = (V0)(t) + (1/2)(g)(t²)
The underlying speed condition is:
V0 = √{(2)(g)(h)}
Substitute √{(2)(g)(h)} for V0:
d = √{(2)(g)(h)}t + (1/2)(g)(t²)
Unravel for h:
√{(2)(g)(h)}t = d - (1/2)(g)(t²)
(2)(g)(h)t² = {d - (1/2)(g)(t²)}²
h = {d - (1/2)(g)(t²)}²/{2(g)(t²)}
h ≈ 0.824 m