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
`        Please provide explanatory answer and reply soon...`
4 months ago

Arun
22984 Points
```							This is a question on horizontal projection. Let R be the horizontal distance covered by the projectile then and a be the angle of projection. U be the initial velocity and in our case it is 16m/sAnd g be the acceleration due to gravity and in our case it is 10.​​Then R = U² Sin 2a / gR = 14.4 + b (it hits the roof half way and the width of the roof is 2b)R = 16² sin 2(45°) / 10= (256 × 0.98769) / 10 = 25.28mb = 25.28 - 14.4 = 10.88m2b = 2 × 10.88 = 21.76mThis is the width of the building. The angle of projection when initial velocity is 10.3m/s and the projectile crosses the roof is given by :Height of the roof equals U Sin a where u is the initial velocity.4.8 = 10.3 × Sin aSin a = 4.8 ÷ 10.3 = 0.4660Sin ⁻¹0. 4660 = 30.86°
```
4 months 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