At highest point vertical component of velocity becomes zero but horizontal components remains.
Net velocity is along horizontal
Also acceleration is vertically downwards throughout the journey
Hence the correct option is (c)
Since velocity is changing, KE and momentum is not constant.
Now since acceleration is vertically downwards, vertical component of velocity is changing.
Since there is no acceleration is horizontal direction, horizontal components is constant
Hence the correct option is (d)
Speed of particle is constant in uniform circular motion
Velocity vector is tangent to the path at any point on the path.
Acceleration vector is directly inwards towards center
So velocity and acceleration vector are perpendicular to each other in Uniform circular motion
Hence the answer is (a) and (b)
R=u2sin2θ2g
So it depends on velocity, angle and acceleration due to gravity
Hence the answer is (a),(b) and (d)
Time taken to reach the ground depends on the velocity in vertical direction...Since both the bodies are projected horizontally. There are no vertical components of velocity involved. So they will reach the ground in same time. Now time taken is same so body with more horizontal velocity will travel more on the ground.
Hence the answer is (a) and (b)
Given
x=2t
y=2t2
Eliminating t we get
y=x2/2
So it is parabola
Vx=dx/dt=2
vy=dy/dt=4t
So velocity at any time t is given by
v=2i+4tj
Now similarly
ax=dVx/dt=0
ay=dVy/dt=4
So acceleration vector is a=4j
Hence Answer is (a),(c) and (d)
Given
u=4i+3j m/s
a=.4i+.3j m/s2
So velocity vector at any time t
v=u+at = 4i+3j+(.4i+.3j)t =(4+.4t)i+(3+.3t)j
So velocity at 10 sec
v=8i+6j
|v|=10
displacement vector at any time t
s=ut+(1/2)at2 =(4i+3j)t+(1/2)(.4i+.3j)t2 =i(4t+.2t2)+j(3t+.15t2)
Hence Answer is (a) and (b)
For projectile A
y=a1x−b1x2
for y=0
x=0 and x=a1b1
Similarly for Projectile B
y=a2x−b2x2
for y=0
x=0 and x=a2b2
For the range to be same
a1b1=a2b2
At the highest point of projectile, vertical components of velocity becomes zero. So there are only horizontal components so, net velocity is horizontal at this point. Now acceleration is vertically downwards throughout the motion. So acceleration and velocity vector are perpendicular to each other at highest point
This question can be solved in two ways
Method 1:
Given
x=36t
y=48t−4.9t2
For a body projected with velocity u at an angle θ with the horizontal ,the x and y displacement is given by
x=(ucosθ)t
y=(usinθ)t−gt22
Comparing this the given equation we have
ucosθ=36
usinθ=48
Squaring and adding we get
u2(cosθ+sin2θ)=3600
so u=60 m/s
Method 2:
Given
x=36t
y=48t−4.9t2
vx=dxdt=36
vy=dydt=48−9.8t
So intial velocity can be found substituting t=0 in both the equation
vx=36
vy=48
So net velocity= √362+482)=60 m/s
If a body is projected with a given velocity u at angle θ and (90−θ) to the horizontal, it will have same range R given by
R=u2sin2θg
The corresponding times if flight are
t1=2usinθg
t2=2usin(90−θ)g=2ucosθg
t1t2=2u2(2sinθcosθ)g2=2u2sin2θg2=2Rg
Range is given by
R=u2sin2θg ----1
and time taken by
T=2usinθg ----2
Now
3R=u2csin2θg ---3
Tc=2ucsinθg ---4
Dividing 1 by 3
13=(uuc)2 ----5
Dividing 2 by 4
TTc=uuc ---6
From 5 and 6
TTc=1√3
Tc=T√3
Range is given by
R=v2sin2θg
Range will maximum when sin2θ=1
So R=v2g
Now area=πR2
=πv4g2
Initial vertical component of velocity is zero
so h=12gt2
39.2=(12×9.8×t2
t=2√2 sec
Now from Projectile motion
h=v20sin2α2g ---(1)
And
r=v20sin2αg ---(2)
Also maximum Range Rm
Rm=v20g ---(3)
Substituting the values of Rm in (1) and (2)
2h=Rmsin2α --(4)
r=Rmsin2α ---(5)
Equation (4) can be rewritten in the form
h=Rm4(1−cos2α)
cos2α=1−4hRm
Equation (5) can be rewritten in the form
sin2α=rRm
Now
sin22α+cos22α=1
(rRm)2+(1−4hRm)2=1
Or, Rm=2h+r28h
It is evident; we must have horizontal velocity of projectile equal to horizontal velocity of the airplane.
So
v0cosα=v ---(1)
If the projectile has to ever attain the height h, then vertical component of projectile must be greater then √2gh
So
v0cosα≥√2gh ---(2)
Squaring equation 1 and 2 .And then adding it
v20≥v2+2gh
v20≥√v2+2gh
(P) The trajectory of bag in case A w.r.t. ground is |
(L) Parabola |
(Q) The trajectory of bag in case B w.r.t. plane is |
(M) straight line |
(R) The trajectory of bag in case B w.r.t. ground is |
(N) no appropriate match |
(S) The trajectory of bag in case A w.r.t. plane is |
(O) Circle |
With a≠0
With respect to plane
dvdt=g−a which is a constant acceleration
Now since v0=0 with respect to plane.
So it is a straight line
With Respect to ground
v≠0 at t=0
Acceleration =g
So it is a parabola
With a=0 and v≠0
With respect to plane
dvdt=g which is a constant acceleration
Now since v0=0 with respect to plane.
So it is a straight line
With Respect to ground
v≠0 at t=0
Acceleration =g
So it is a parabola
Column A |
Column B |
A) tan of angle between velocity and x axis at any point of time |
|
B) tan of projection angle θ0 for a projectile at origin |
|
C) Range of the projectile |
|
D) tan of the angle of position vector at any time |
|
|
(A) (T);(B) (S);(C) (P);(D) (Q)
At maximum height ,Y components of velocity becomes zero, So X components remains only
So v=32vi
Total time when the stone hit X-axis =2 × Time to reach maximum height
T=2vg√32
or
T=v√3g
Distance travelled along X -axis in this time will be given as
=32v×v√3g=3√32v2g
So, co-ordinates are L+3√32v2g,0
When the stone hit the X-axis, The horizontal velocity will remain same while the vertical components of the velocity will be downward direction
32vi−frac√32vj
Total time
T=v√3g
So distance travelled =v2√3g
So co-oridinates are
L+v2√3g,0
Consider the figure given below and writing the equation for point X
x=(v0cosθ)t=Rcosα - (1)
y=(v0sinθ)t−12gt2=Rsinα -(2)
Eliminating t between equation (1) and (2),we get
Now for R to be maximum
sin(2θ−α)=1
tan2θ+cotα=0
Let us just assume that both the outer walls are equal in height say h and they are at equal distance x from the end points of the parabolic trajectory as can be shown below in the figure.
Now equation of the parabola is
y=bx−cx2-- (1)
y=0 at x=nr=R
where R is the range of the parabola.
Putting these values in equation (1) we get
b=cnr -- (2)
Now the range R of the parabola is
R=a+r+2r+a=nr
This gives
a=(n−3)r2 -- (3)
The trajectory of the stone passes through the top of the three walls whose coordinates are
(a,h),(a+r,157h),(a+3r,h)
Using these co-ordinates in equation 1 we get
h=ab−ca2 -- (4)
157h=b(a+r)−c(a+r)2 --(5)
h=b(a+3r)−c(a+3r)2 ----(6)
After combining (2), (3), (4), (5) and (6) and solving them we get n = 4.