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
Menu
Grade: 9
        
A ball is thrown vertically upwards from the top of a tower at 4.9 ms. it strikes the pond near the base of the tower after 3 seconds The height of the is Take g = 9.8 m/s
one month ago

Answers : (1)

Sai Ashish Vure
26 Points
							
Using s = ut + (1/2)at^2, with downwards as positive,
s = (-4.9m/s * 3s) + (1/2) * 9.8m/s^2 * (3s)^2
s = -14.7m + 4.9m/s^2 * 9s^2
s = -14.7m + 44.1m
s = 29.4m
Hence, the height of the tower is 29.4m.
one month ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details