 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
`        A body of mass 100 gram is projected vertically up with a velocity of 10 metre per second and another body of same mass is projected vertically down with the same velocity from the top of the tower of height 20 metre calculate the kinetic energy of each ball on reaching the ground take g is equal to 10 metre per second square`
5 months ago

```							 Let the distance covered by the 100g ball before collision be ‘x’.s = ut+[(1/2)at*t]s = x, u = 49m/s, a = – 9.8m/(s*s),  t = ?.Therefore,  x = 49t + [(1/2)*( – 9.8)*t*t]           ...(i)The height of the tower is 98m so, the distance covered by the other ball is 98 – xHere s = 98 – x,  u = 0, a = 9.8m/(s*s), t = ? (equal to the time of flight of the other ball)so, 98 – x = 0t + [(1/2)*(9.8)*t*t]            x = 98 – [(1/2)*(9.8)*t*t]from (i) we get x = 49t – [(1/2)*(9.8)*t*t]Therefore 98=49tand then t = 2s.and x = 49*2 – [(1/2)*(9.8)*4]x = 78.4m.Now after collisions = 78.4m,  u = 0,  a = 9.8m/(s*s),  t = ?.Therefore 78.4 = ½ (9.8)(t*t)t*t = (78.4)*2 / 9.8t*t = 16t = 4. Therefore total time of flight is 6s
```
5 months ago Saurabh Koranglekar
3154 Points
``` ```
5 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »  ### 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