×

#### 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
```
an elevator without a ceiling is ascending with a constant speed of 10m/s. a boy on the elevator shoots a ball directly upward, from a height of 2.0 m above the elevator floor. At this time the elevator floor is 28 m above the ground. the initial speed of the ball with respect to the elevator is 20 m/s (take g=9.8m/s²) a). what maximum height above the ground does the ball reach? b). How long does it take to return to the elevator floor

```
2 years ago

2057 Points
```							this is a question of ncert.	the initial speed of ball is 20+10=30m/s. so, u=30, v=0, a= –9.8. 2*-9.8*h= – 900. so, max height is 30+45.9=75.9m or approx 76m.	taking the reference frame as the elevator, we have u=20, a= –9.8, v=0. so, 0=20-9.8t or t=2.04. so total time taken to return to the elevator floor is 4.08 seconds
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Trigonometry

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

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