Question

pls tell que no. 6.........

Answer 1

I am solving other questions also.  They are interesting too...

Qn 6:

      Assume frictionless ground.

     Let us take the point of projection as the origin.  The horizontal right is x axis and the vertical direction upwards is y axis.    Let g = 10 m/s².
 

    When the projectile meets the inclined slope of the wedge, the (x,y) of the wedge and projectile are same at a "t".

Horizontal distance travelled by the projectile :
      x = u cos 60° * t = 5 √3 t     --- (1)

Equation of the inclined plane of the wedge: y = -x/√3  meters
As the wedge moves the equation is:       y = (10√3 t - x) /√3  = 10 t - x/√3
Substitute for x from  (1):     y = 10 t - 5 t = 5 t    --- (2)

Equation of the projectile:  y = x tan 60° - g x² sec² 60°/2u²
            =>         y =  √3 x - x²/15

Substituting for x from (1) and (2):     
         5 t = √3 * 5 √3 t  - (5 √3 t)² / 15
           t² = 2 t
        So either t = 0   or  2 sec.      Answer:  2 seconds


 Question 7:

   R = range of a projectile = u² Sin 2θ / g = 80*80 Sin 2θ /32 = 200 sin 2θ ft

   R = 100 ft = 200 sin 2θ ft =>
   sin 2θ = 1/2 => θ = 15 deg or 75 deg.

Longest time taken for a projectile to reach 100 ft: 100 / (80 cos 75) = 4.83 s
Shortest time taken for a projectile to reach 100 ft: 100/ (80 cos 15) = 1.29 s

The time interval between the two = 4.83 - 1.29 = 3.54 sec = 5/√2 sec.

If you use cos 15 = (√3+1)/2√2 and cos 75 = (√3 -1)/2√2 
Then the interval you will get exactly as 5/√2 sec.
 

Qn 8: Balloon problem

y = V0* t           Vx = a y = a Vo* t
           so x = (aVo/2) t^2                      y^2 = (2 V0/a) x

dy/dx = sqrt(V0/ax) ;;; d^2y/dx^2 = - sqrt[V0/8ax^3]

R = radius of curvature = [(2ax+V0)^3/2] / [a sqrt(V0) ]

velocity V^2 = (dx/dt)^2 + (dy/dt)^2 

V = V0 sqrt[1+a^2 t^2] = sqrt[V0^2 + 2aV0 * x ]

(normal) radial acceleration = V^2/R = aV0 / SQRT[ 1 + 2a x /V0 ]

d^2 y/dt^2 = 0 ;; d^2x/dt^2 = aV0.

slope = tan theta = dy/dx = sqrt[V0/2ax] ;;; Cos theta = sqrt[[2ax/(V0+2ax) ]
tangential acceleration = d^2x / dt^2 * Cos theta = aV0 * sqrt[2ax/(V0+2a x) ]
verify by sum of squares of two accelerations = [a V0]^2

Qn 5 : Interesting one.... on Drag force..

m d^2 y/dt^2 = mg – b dy/dt … ---- (1)
m dy/dt = m g t – b y

solution is of the form : y = A t + B + C e^(-Dt) 
       so  y’ = A – CD e^(-D t)

Substitute, and find constants: y = (gm/b) t -(gm^2/b^2)[1 - e^(-b t /m)] 
         Vy = (mg/b) [1+e^(-bt/m)]

Similarly, m d^2x/dt^2 = 0 – b dx/dtm dx/dt – m V0 = - b x

Integrating:    x = (mV0/b) [ 1 – e^(-bt/m) ]

Vx = V0 e^(-bt/m)

Instantaneous velocity = V = sqrt(Vx^2 + Vy^2)