badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

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

Please check your email for login details.
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: 12th pass

                        

a ball is thrown so that it just clears a 3 meter fence 18 meters away. If it left the hand 1.5 meters above the ground and at an angle of 60 o with the horizontal, what was the initial velocity of the ball?

3 years ago

Answers : (1)

Sounak Dutta
27 Points
							
Resolving into components we will find that horizontal velocity is v*cos60whereas vertical component is v*sin600....(we are assuming the velocity of the ball as v initially)
So...forming equations let the ball reach the wall in time t.
     so along horizontal
                     v*cos600*t=18             as distance between thrower and wall is 18 metres.
    Along vertical,
                 using formula v2=u2-2*a*s..
      here u being v*sin600 and v being 0,,a being g,i.e., 9.8m/s
        s = 1.5m
 we get another equation in v...
 but...by only solving second equation....we get our required velocity as 39.2 m/s
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

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