Kinematics

R = (2 u² Sin θ Cos θ)/gH = (u Sin θ)²/ 2gT = (2 u Sin θ)/g(Time of flight = t)
Now when wind blows in horizontal direction
R (new) = u Cos θ*t + 1/2 (g/2) t² (Applying equation of motion)
R (new) = (2 u² Sin θ Cos θ)/g + (g/4) (4 u² Sin²θ)/g²
Solving we get
R (new) = R + (u² Sin²θ)/g
R (new) = R + 2H (Answer)

Now, when the aircraft will drop the food package, it will obviously travel a parabolic path.
Let the aircraft be flying with the horizontal speed = v m·s−1.
So, the horizontal velocity given to the food package when left will be same = v m·s−1.
The vertical component will be zero i.e. it is coming down by the effect of gravity.
Let us say the vertical height (altitude) of the aircraft = x m.
Let us say the total time taken by the package to reach the ground = t seconds.
So, Initial vertical downward velocity u = 0 m·s−1.
Distance = -x  ( Vertical Downward displacement)
Acceleration due to gravity (g) = -10 m·s−2
Time = t seconds
So -x = 1/2 (-10) t²
Cancelling the negative sign, x = 5 *  t²
Now in last 6 seconds, package covered the distance of 780 m
So, distance covered in (t-6) seconds = (x - 780) m
Now, again applying the equation-(x-780) = 1/2 (-10) (t-6)²
Cancelling the negative sign, and solving we get
 t = 16 seconds
Putting in first equation x = 5 * t²
we get x = 1280 m = 1.28 km (Answer)
Since, it has travelled a horizontal distance of 1000 m before 6 seconds, so this distance has been covered in (16 - 6 = 10 s)
Distance = Speed * Time
1000 = v * 10 (Horizontal component given to food package is same as speed of aircraft)
v = 100 m·s−1 = 360 km/ hr (Answer).

We can use vectors for solving the above problem
Now horizontal velocity remains the same = v Cos θ
Vertical Initial Velocity = + v Sin θ
After time t, instantaneous horizontal velocity = v Cos θ (same throughout)
Instantaneous Vertical component
v = u + atv = v Sinθ - gt (-g because g acts downwards)
Instantaneous velocity vector (V)V = v Cos θ î + (v Sinθ - gt)
Average velocity = Total Displacement/ Total Time
Displacement in X direction = v Cos θ * t
 Displacement in Y direction = v Sin θ* t  - gt²
Writing in vector form, displacement vector= (v Cos θ * t )î + (v Sin θ* t - gt²)
Dividing by time t (total time from 0 to t)
Average velocity vector AV1 =  (v Cos θ)î + (v Sin θ - gt)
Taking the magnitude of AV1 and V, and solving
we get
t = (4 v Sin θ)/(3 g)

Differentiating R with respect to time, we get V = (2t - 4) î + (2t) ĵ
Again differentiating velocity with respect to time, we get A= (2) î + (2) ĵ
Since velocity and acceleration vector are perpendicular to each other, their Dot product should be
0AV =  (2t - 4) î + (2t) ĵ .  (2) î + (2) ĵ = 04t - 8 +4t = 08t = 8t =1 s (Answer)

Let the horizontal component of the speed of the bullet when it passes through A is v m•s−1.
Now it's vertical component is 0m•s−1.
It falls down a height of 0.1m (10 cm).
Now time taken to travel this vertical distance, Vertical distance = (-0.1 m)
Initial vertical component (u) = 0 m•s−1
S = ut - 1/2(gt²)
Now S = -1/2 (gt²) -0.1 = -4.9 * t²
Cancelling the negative sign
t = 1/7 seconds
Now horizontally time taken by the bullet to travel the horizontal distance is same
Distance = SpeedTime
100 = v * (1/7)v = 700 m•s−1

Differentiating y with respect to time and putting it to 0,
10 - 2t = 0t = 5 s
Now after 5 s, it will attain maximum height as (dy/dt = v =0) at maximum height
(Vertical component)putting value of t = 5 s in y, 
y = 50 -25 = 25 m (Correct Answer)

Since the balloon is going upwards with a constant speed of +25 m•s−1 (+ is a sign convention for upward speed).
After 5 seconds, it will be at a height of +125 m (25 m•s−1 X 5 s).
Now after 5 seconds let say, the bullet is shot with a speed of +u m•s−1.
Applying the concept of relative velocity, we stop the balloon at a height of +125 m.
We reverses its speed = (-25 m•s−1) and add into the speed of the bullet.
Now speed of bullet becomes, (u - 25) m•s−1
The distance, bullet has to travel = +125 m
Applying the equation,v2-u2 = 2gS , v is the final speed of the bullet = 0 m•s−1 (As it just reaches the balloon and stops),
Initial Speed (u) = (u-25) m•s−1, S=+125 m, g = -10 m•s−2 (- because g always acts in the downward direction)
Solving the equation,(0)2- (u-25)2 = 2 X (-10)X (+125)-(u-25)2 = - (2500)
Cancelling the negative sign, and taking the square root, we get u-25 = 50 u = 75 m•s−1 (Answer)

Applying the equation v² - u² = 2aS
0² - (20)² = 2a2
-400 = 4a
a = -100 km/ hr
-2v = u + at
0 = 20 - 100*t
t = 0.2
Distance travelled by the bird = 0.2 * 60 = 12 km (Answer)

Initial vertical component = 0 m.s^-1
Distance to be traveled = -h
Applying the second equation of motion,
-h = 1/2(-g)(t²)
t = √(2h/g)
Horizontal component of velocity = √2gh
Distance = velocity* Time
x = √2gh*√(2h/g) = 2h (Correct Answer)

Let the total height till bottom of the window be x m.
Then, the height from the top of the building to the top of the window = (x - 3) m .
Let time taken by the ball to reach from top of the building to the top of the window = t s,
then the time taken to reach the bottom of the window from the top of the building is (t + 0.5) s.
Applying the equation,
(x - 3) = 5t²
x = 5 (t + 0.5)²
Solving, we get
t = 0.35 seconds
x = 3.6125 m
Applying the third equation of motion
u = 0 m.s-1
Vt² - u² = 2g0.6125
Solving we get Vt = 3.5 m.s-1
Now again applying the third equation of motion,
Vb² - u² = 2g*3.6125
Solving we get Vb = 8.5 m.s-1
Adding the two
Vt + Vb = 12 m.s-1 (Correct Answer)

 Differentiating the equation with respect to time,
1 = α (2x) (dx/dt) + β(dx/dt)
dx/dt = v
1 = α (2x) (v) + β(v)----------- (1)
x = (1- β(v))/(2 α (v))----------- (2)
Again differentiating (1) with respect to time
0 = α (2x) (dv/dt) + (2αv)(dx/dt) + β (dv/dt)
dv/dt = A (retardation)
0 = α (2x) (A) + (2αv)(v) + β (A)
Substituting x in terms of v as in (2)
we get A = -2αv³ 

Taking ball 1, Velocity along let say x direction = 10√3 m.s^-1
Velocity along y direction = 30 m.s^-1 
Resultant velocity is 20√3 m.s^-1 at an angle of 60 degrees with x direction
(tan x = 30/10√3)
x = 60 degrees
Similarly for second ball resultant is = 20√2 m.s^-1 at an angle of 45 degrees.
(tan x = 20/20 = 1)
x =45 degrees
So, the angle between their directions of motion is 60 - 45 = 15 degrees (Correct Answer)

Velocity Vector is given by the unit vector in that direction and speed
Initial Point (1, -2, 3)
In Vector Form: (i - 2j + 3k)
Final Point (4, 2, 3)
In Vector Form: (4i + 2j + 3k)
Position Vector can be represented by(4-1)i + (2-(-2))j + (3-3)k(3)i + (4)j----------------(1)
Magnitude = (3²+4²)^1/2
Magnitude = 5-----------(2)
Unit Vector = Dividing (1) by (2)
we get 0.6i + 0.8j
Now speed of the bird = 10 m.s-1
Vector is represented by magnitude and direction
Now magnitude of velocity is speed
So 10 (0.6i + 0.8j) = 6i + 8j (Correct Answer)

 The stone is dropped into the well. So it travels a distance of h, and after falling into the water, sound comes out of the well after travelling distance h with velocity v.
Time (t₁) taken by the stone to travel the height h down the well
Applying the equation,
-h = (1/2)(-g)(t₁)²t₁ = (2h/g)^1/2
Now after falling into the water, sound travels the distance h upwards with speed v
Time taken by the sound to come out of the well t₂
t₂ = (h/v) (Time = Distance/ Speed)
Total time taken T = t₁ +t₂
T = (2h/g)^1/2 + (h/v) (Correct Answer)

Let say the velocity of boat 1 with respect to boat 2 is V₁₂
V₁₂ = V₁ - V₂
It is given that the direction is 60° north of east.
Let towards east we have (+i) (unit vector i cap)
In north direction we have (+j) (unit vector j cap)
So in south we will have (-j)
East component of velocity = 10√3 Cos60° = 5√3
North component of velocity = 10√3 Sin60° = 15
Thus V₁₂= 5√3i + 15j (In vector form)
V₂ = -15j (It is in the south direction that's why it is negative)
V₁₂ = V₁ - V₂5√3i + 15j  
        = V₁ - (-15j)5√3i + 15j
        = V₁ +15j
So  V₁ = 5√3i (i.e. 5√3 km.hr-1 in the east direction) (Correct Answer)