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 = u*t - 1/2(g*t²) Now S = -1/2 (g*t²) -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 = Speed*Time 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 = 2*a*S 0² - (20)² = 2*a*2 -400 = 4*a a = -100 km/ hr -2v = u + a*t 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) = 5*t² 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 = 2*g*0.6125 Solving we get Vt = 3.5 m.s-1 Now again applying the third equation of motion, Vb² - u² = 2*g*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)