Click to Chat

1800-2000-838

+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: R

There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

• Complete Physics Course - Class 11
• OFFERED PRICE: R 2,800
• View Details
Get extra R 560 off
USE CODE: MOB20


Two Particles A & B are projected from the same point with the same velocity u of projection but at different angles (x) & (y), such that the maximum height of A is two-third of the horizontal range of B.Then which of the following relations are true ?
(a) Range of A = Maximum height of B
(b) Maximum horizontal range of A = u^2/g and this occurs when (y) = 1/2 sin^-1 3/8
(c) 3(1 - cos 2{x}) = 8 sin 2(y)
(d) Maximum value of (y) is 1/2 sin^-1 3/4

[Ans: (b) , (c) , (d)]



6 years ago

Share

										Dear   zahid sharief,
The initial velocity v0
$v_0=v_{0x}\mathbf{i} + v_{0y}\mathbf{j}$
if the angle θ0 is known:
v0x = v0cosθ0 v0y = v0sinθ0

The horizontal displacement(x − x0 )

$x-x_0=v_{0x}t+\begin{matrix}\frac{1}{2}\end{matrix}at^2$

x − x0 = (v0cosθ0)t.
.$y-y_0=(v_0\sin\theta_0)t-\begin{matrix}\frac{1}{2}\end{matrix}gt^2$.

$y-y_0=(v_0\sin\theta_0)t-\begin{matrix}\frac{1}{2}\end{matrix}gt^2$

equation of projectile
$y=x\tan(\theta)-x^2g/(2v^2_{0}cos^2(\theta))$
Time to reach the maximum height
$t={v_{0y}\over g}$
Range of projectile
${r=\frac{v^2}{g}\sin(2\theta)}\,$
Maximum height
$y_{max}={{v^2\sin^2\theta}\over 2g}$

maimum height of  A     y1(max) =   v^2 (1-cos 2x)/4 *g       here  2 sin2x= 1-cos2x
range of                        r2(max) = v^2*sin(2y)/g

compare them we get    3(1 - cos 2{x}) = 8 sin 2(y)

maimum value    (y)= 1/2 [sin-1 {3/8(1 - cos 2{x}) } ]

say cos2x=-1
we get maximum value   that is     max(y)= 1/2 [sin-1 {3*2/8} ]
(y)= 1/2 [sin-1 3/4 ]

All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Rathod Shankar
IIT bombay

6 years ago

Other Related Questions on Mechanics

a packet is released from a balloon accelerating upward with acceleration a. the acceleration of packet just after the release is

Since the ballon was moving upward with acceleration a, there is a force of gravity on the packet too. The balloon is moving upward bexause the reaction force R>W, weight of the packet. As...

 Shaswata Biswas 4 months ago

Just after the release the packet posses only the component of the velocity aquired. Here the accleration of the balloon has no effect on the acceration of the body. When the body is...

 Shaswata Biswas 4 months ago

The accleration of the packet will be g acting downward as no external force other then gravity working on it.it will achieve the case of free fall. If I am wrong please ..explain the...

 Chandan kumar mandal 2 months ago

Since the object is in the influence of both the accelerations initially, It's acceleration just after release will be the vector sum of acceleration due to gravity and acceleration of the...

 yesterday

acceleration due to gravity =g.Because as soon as the packet droped from ballon,it looses contact of balloon.It is under free fall now.There fore it occurs the acceleration due to gravity.

 yesterday
The potential energy of a particle of mass 1kg moving along x-axis is given by U(x)=[x^2/2-x]J. If total mechanical energy of the particle is 2J find its maximum speed.

total mechanical energy = U+ K.E to attain maximum speed the obect must have maximum K.E as K.E will be maximum , U has to be minimum (conservation of energy )given - U(x) = x²/2-xfor this...

 fizaparveen 2 months ago

We know that the object accelerate till force is applied on it and it attains maximum velocity just after the force becomes zero... SO we know the negative of potential energy gradient is...

 Suraj Singh 26 days ago

Ans- 2Speed will be max when kinetic energy is maximum so potential energy will be minimum so differentiating the function put it equal to 0 so we obtain minimum for x=0 now kin potential...

 rishabh doshi 2 months ago
if a body move with u velocity then after collide with wall and remove with v velocity what is the impulse

Hi aman, I think this might help as we know that Impulse and also F=m(dv/dt) so (here m is the mass of body and dv/dt =acc) here p is the momentum ans delta p is change in momentum hence in...

 Ankit Jaiswal 3 months ago
what is work energy thereom...................................?

The work-energy theorem is a generalized description of motion which states that work done by the sum of all forces acting on an object is equal to change in that object’s kinetic energy....

 dolly bhatia one month ago

The work W done by the net force on a particle equals the change in the particles kinetic energy KE: W = Δ K E = 1 2 m v f 2 − 1 2 m v i 2 . The work-energy theorem can be derived from...

 Sushmit Trivedi one month ago

For any net force acting on a particle moving along any curvilinear path, it can be demonstrated that its work equals the change in the kinetic energy of the particle by a simple derivation...

 rahul kushwaha 3 months ago
The cieling of long hall is 25 m high What is the maximum horizontal distance that a ball thrown with a speed of 40 ms can go without hitting the ceiling of wall? Plz explain with a DIAGRAM

Let the angle of throwing be Q with respect to the ground So, the range is R = 40*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...

 Tapas Khanda 5 months ago

Let the angle of throwing be Q with respect to the ground So, the range is R = 30*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...

 Tapas Khanda 5 months ago

Diagram is not required to solve this question. You can directly use the formula to find range : R = (v^2/g)*sin2Q Maximum height reached, H = v^2*sin^2Q/2g So, 25 = 40^2*sin^2Q/(2*9.8) So,...

 Tapas Khanda 5 months ago
What harmfull gases are produced when we burn plastics? And can we burn plastics?

Im doing a project, can someone help me, How to reduce those toxic contents when plantic are burnt Like sulphur, nitrogen etc

 2017 years ago

nitrogen gas , sulphur dioxide ,carbon monoxide , carbon dioxide. are some of the gases. well burning plastic is not good for living things. but there are many other things that are more...

 4 days ago
View all Questions »

• Complete Physics Course - Class 12
• OFFERED PRICE: R 2,600
• View Details
Get extra R 520 off
USE CODE: MOB20

• Complete Physics Course - Class 11
• OFFERED PRICE: R 2,800
• View Details

Get extra R 560 off
USE CODE: MOB20

More Questions On Mechanics