 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
`        AT WHAT ANGLE WITH THE HORIZONTAL SHOULD A BALL BE THRWN SO THAT THE RANGE R IS RELATED TO TIME OF FLIGHT ‘T’ AS R=5T^2`
2 years ago

```							R=5T^2(u^2*sin2 theta)/g=5*[(2*u*sin theta)/g]^2(u^2*sin2 theta)/g=5*4*u^2*sin^ 2theta/g^2sin2 theta*g=5*4*sin^2theta10sin2 theta=20sin^2 theta (g=10)sin2 theta=2sin^2 theta2sin theta*cos theta=2sin^2 theta  (sin2 theta=2sin theta*cos theta)sin theta*cos theta=sin^2 thetacos theta = sin thetacot theta = 1i.e. theta= 45 degree
```
2 years 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