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)
$R=\frac {u^2sin 2 \theta}{2g}$
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=2t^{2}
Eliminating t we get
y=x^{2}/2
So it is parabola
V_{x}=dx/dt=2
v_{y}=dy/dt=4t
So velocity at any time t is given by
v=2i+4tj
Now similarly
a_{x}=dV_{x}/dt=0
a_{y}=dV_{y}/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/s^{2}
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)at^{2} =(4i+3j)t+(1/2)(.4i+.3j)t^{2} =i(4t+.2t^{2})+j(3t+.15t^{2})
Hence Answer is (a) and (b)
For projectile A
$y=a_1xb_1x^2$
for y=0
$x=0$ and $x=\frac {a_1}{b_1}$
Similarly for Projectile B
$y=a_2xb_2x^2$
for y=0
$x=0$ and $x=\frac {a_2}{b_2}$
For the range to be same
$\frac {a_1}{b_1} =\frac {a_2}{b_2} $
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
$x=36t$
$y=48t4.9t^2$
For a body projected with velocity u at an angle $ \theta$ with the horizontal ,the x and y displacement is given by
$x=(ucos \theta)t$
$y=(usin \theta )t\frac {gt^2}{2}$
Comparing this the given equation we have
$ucos \theta=36$
$usin \theta=48$
Squaring and adding we get
$u^2(cos^ \theta +sin^2 \theta)=3600$
so u=60 m/s
Method 2:
Given
$x=36t$
$y=48t4.9t^2$
$v_x=\frac {dx}{dt}=36$
$v_y=\frac {dy}{dt}=489.8t$
So intial velocity can be found substituting t=0 in both the equation
$v_x=36$
$v_y=48$
So net velocity= $\sqrt {362+482)}=60 $ m/s
If a body is projected with a given velocity u at angle $\theta$ and $(90 \theta)$ to the horizontal, it will have same range R given by
$ R=\frac {u^2 sin2 \theta}{g}$
The corresponding times if flight are
$t_1=\frac {2usin \theta}{g}$
$t_2=\frac {2usin(90 \theta)}{g}= \frac {2ucos \theta}{g}$
$t_1t_2=\frac {2u^2(2sin \theta cos \theta)}{g^2} =\frac {2u^2 sin2 \theta} {g^2}=\frac {2R}{g}$
Range is given by
$R=\frac {u^2 sin 2 \theta}{g}$ 1
and time taken by
$T=\frac {2usin \theta}{g}$ 2
Now
$3R=\frac {u_c^2 sin2 \theta }{g}$ 3
$T_c=\frac {2u_c sin \theta }{g}$ 4
Dividing 1 by 3
$ \frac {1}{3}=(\frac {u}{u_c})^2$ 5
Dividing 2 by 4
$\frac {T}{T_c}=\frac {u}{u_c}$ 6
From 5 and 6
$\frac {T}{T_c}=\frac {1}{ \sqrt {3}}$
$T_c=\frac {T}{ \sqrt {3}}$
Range is given by
$R=\frac {v^2 sin2 \theta }{g}$
Range will maximum when $sin2 \theta =1$
So $R= \frac {v^2}{g}$
Now $area= \pi R^2$
$=\frac { \pi v^4}{g^2}$
Initial vertical component of velocity is zero
so $h=\frac {1}{2}gt^2$
$39.2=(\frac {1}{2} \times 9.8 \times t^2$
$t=2 \sqrt {2}$ sec
Now from Projectile motion
$h= \frac {v_0^2 sin^2 \alpha}{2g}$ (1)
And
$r= \frac {v_0^2 sin 2 \alpha}{g}$ (2)
Also maximum Range $R_m$
$R_m = \frac {v_0^2}{g}$ (3)
Substituting the values of $R_m$ in (1) and (2)
$2h = R_m sin^2 \alpha$ (4)
$r = R_m sin2 \alpha$ (5)
Equation (4) can be rewritten in the form
$h = \frac {R_m}{4}(1  cos 2 \alpha)$
$ cos 2 \alpha = 1  \frac {4h}{R_m}$
Equation (5) can be rewritten in the form
$ sin 2 \alpha = \frac {r}{R_m}$
Now
$ sin^ 2 2\alpha + cos ^2 2\alpha=1$
$ (\frac {r}{R_m})^2 + (1  \frac {4h}{R_m})^2 =1$
Or, $R_m=2h + \frac {r^2}{8h}$
It is evident; we must have horizontal velocity of projectile equal to horizontal velocity of the airplane.
So
$v_0 cos \alpha = v$ (1)
If the projectile has to ever attain the height h, then vertical component of projectile must be greater then $\sqrt {2gh}$
So
$v_0 cos \alpha \geq \sqrt {2gh}$ (2)
Squaring equation 1 and 2 .And then adding it
$v_0^2 \geq v^2 + 2gh$
$v_0^2 \geq \sqrt { v^2 + 2gh}$
(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 \neq 0$
With respect to plane
$ \frac {d\mathbf{v}}{dt} = \mathbf{g} \mathbf{a}$ which is a constant acceleration
Now since $v_0=0$ with respect to plane.
So it is a straight line
With Respect to ground
$v \neq 0$ at t=0
Acceleration =$\mathbf{g}$
So it is a parabola
With a=0 and $v \neq 0$
With respect to plane
$ \frac {d \mathbf{v}}{dt} =\mathbf{g}$ which is a constant acceleration
Now since $v_0=0$ with respect to plane.
So it is a straight line
With Respect to ground
$v \neq 0$ 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 θ_{0} 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 $\mathbf{v} =\frac {3}{2}v \mathbf{i}$
Total time when the stone hit Xaxis =2 × Time to reach maximum height
$ T = \frac {2v}{g} \frac {\sqrt {3}}{2}$
or
$T= \frac {v \sqrt {3}}{g}$
Distance travelled along X axis in this time will be given as
$=\frac {3}{2}v \times \frac {v \sqrt {3}}{g} = \frac {3 \sqrt {3}}{2} \frac {v^2}{g}$
So, coordinates are $ L + \frac {3 \sqrt {3}}{2} \frac {v^2}{g}, 0$
When the stone hit the Xaxis, The horizontal velocity will remain same while the vertical components of the velocity will be downward direction
$ \frac {3}{2}v \mathbf{i}  frac {\sqrt {3}}{2}v \mathbf{j}$
Total time
$T= \frac {v \sqrt {3}}{g}$
So distance travelled $= \frac {v^2 \sqrt {3}}{g}$
So cooridinates are
$L + \frac {v^2 \sqrt {3}}{g}, 0$
Consider the figure given below and writing the equation for point X
$x = (v_0 cos \theta) t = R cos \alpha$  (1)
$y = (v_0 sin \theta)t  \frac {1}{2} gt^2= R sin \alpha $ (2)
Eliminating t between equation (1) and (2),we get
Now for R to be maximum
$ sin (2 \theta  \alpha)=1$
$ tan 2 \theta + cot \alpha=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
\(y = bx  c{x^2}\) (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
\(a = \left( {n  3} \right)\frac{r}{2}\)  (3)
The trajectory of the stone passes through the top of the three walls whose coordinates are
\(\left( {a,h} \right),\left( {a + r,\frac{{15}}{7}h} \right),\left( {a + 3r,h} \right)\)
Using these coordinates in equation 1 we get
\(h = ab  c{a^2}\)  (4)
\(\frac{{15}}{7}h = b(a + r)  c{(a + r)^2}\) (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.