×

#### 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
```
The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of motion of the projectile, giving it a constant horizontal acceleration= g/2. Under the same conditions of projectile, find the horizontal range of the projectile
The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of motion of the projectile, giving it a constant horizontal acceleration= g/2. Under the same conditions of projectile, find the horizontal range of the projectile

```
5 years ago

mohit vashishtha
144 Points
```							new range will be R+2HAPPROVE IF USEFUL
```
5 years ago
AP
12 Points
```							How? Will you please explain?
```
5 years ago
Hasan Naqvi
97 Points
```							Time taken by projective to reach ground = 2 * (2H/g)1/2S= ut + Β½ at2= R + Β½ * g/2 *2* ((2H/g)1/2)2=R + g * H/g=R+ HΒ
```
5 years ago
Nicho priyatham
625 Points
```							R+2HΒ we know that in 1st Β caseΒ Β H=(u2sin2a)/2g Β  R=(u2sin2a)/gnow wright the equation of motion in second cases=((ucosa)t-(1/2)(g/2)t2Β )i-((usina)t-(1/2)gt2Β )jwhen partical lands on ground coffeicant of jΒ =0 so t=(2usina)/g subatute above equation in cofficant of iΒ to Β get new rangeso we getΒ (u2sin2a)/g +(u2sin2a)/g=R+2HΒ
```
5 years ago
kritika sharma
11 Points
```							so its total time of flight and Hmax will be same T=2Vy/g & Hmax=Vy^2/2gbecause of the wind horizontal acc. will be g so range will changeRnew=VxT+1/2 gT^2 =2VxVy/g + 2Vy^2/gso, 2VxVy/g is normal range & 2Vy^2/g=4Hso ans. is R+4H.Β
```
3 years ago
Raahul
31 Points
```							πππππππππππππππππππππππππππππππππππππππππππ€£π­πππππ€ππ­ππΆπ€£ππππππππ€πππ­πππππ€π
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions Β»

### 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

Post Question