×

Thank you for registering.

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

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
```
Two balls are projected upward simultaneously with speed 40m/s and 60m/s . Relative position of second ball w.r.t first ball at time t=5s is

```
5 years ago

Hasan Naqvi
97 Points
```							At t=5s;Distance covered by first ball = 40*5 – ½ * 10 * 52 = 200 – 125 = 75mDistance covered by second ball= 60 * 5 – ½ * 10 * 52 = 300 – 125 = 175m Relative position of second ball wrt to first ball = 175-75 =100m
```
5 years ago
shubham
15 Points
```							Range = velocity × time R1=40×5=200R2=60×5=300relative position is R2-R1 so 300-200=100 100 m is the  correct answer....
```
2 years 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