Questions - Answers
 

1. Derive s = ut +   at² graphically.

A: A body having initial velocity 'u' moves with uniform acceleration 'a'. Its velocity becomes '' after a time 't'. It makes displacement 's' during that time. A graph plotted between velocity and time for this body is a straight line with Y-intercept 'u' and slope 'a'.

Area between  - t graph and time axis gives the displacement of the body. 

  s = area of the trapezium ABCD =   (Sum of the parallel sides) × Perpendicular distance between them

              =  (u + )t

              =   (u + u + at)t

              = ut +  at². 

2. Derive s_n = u + a(n -

 ).

A: Initial velocity of a body is 'u'. When it moves with uniform acceleration 'a' its final velocity is  after a time t. To find its displacement in nth second.

          Displacement during nth second = displacement in n seconds - displacement in (n - 1) seconds.

                

3. Show that the time of ascent is equal to time of descent for a body projected vertically upwards.

A: Time of ascent is the time taken by a body to reach its maximum height position where its final velocity is zero. So, substituting   = 0 in   = u - gt, one gets u = gt_a. Time of ascent t_a = u/g. If the maximum height reached is h_max i.e., y = h_max substituting in

                 y = ut -  gt² h_max = ut_a -   gt_a² = (gt_a) t_a -   gt_a² =  gt_a²

                                

    The body on reaching its maximum height position falls freely with initial velocity u = 0. The time taken by it to reach the ground is time of descent (t_d). From y =   gt²

                             h_max = 

  gt_d²   =

   Hence time of ascent = time of descent.
 

4. Show that the path of a projectile is a parabola.

A: A body thrown at an angle other than 0° or 90° with the horizontal is an oblique projectile. If the initial velocity of the projectile is u. It is thrown at an angle θ with the horizontal. Its horizontal and vertical components are u cos θ and u sin θ respectively. Initially i.e., at t = 0 the projectile is at the origin. At time t its position coordinates are (x, y). In time t = t, vertical displacement s = y, vertical acceleration a = -g, vertical initial velocity = u sin θ. Substituting in

s = ut +   at² the vertical displacement y = (u sin θ) t -   gt²

Horizontal displacement s = x,

Horizontal acceleration a = 0,

neglecting air resistance horizontal initial velocity 

             u = u cos θ

Again using s = ut +   at² the horizontal displacement

             x = ( u cos θ ) t.

Substituting the value of t = x/u cos θ in

          y = (u sin θ ) t -   gt² 

Where A = tan θ and B = g/2u² cos² θ

 This is the equation of parabola. Hence the path of the projectile is a parabola.

5. Obtain expressions for time of flight, maximum height and horizontal range of an oblique projectile.

A: Time of flight is the time taken by a projectile to reach the horizontal level of projection. Using the equation of vertical displacement y = (u sin θ)t-   gt² and substituting y = 0 and t = T

                        We get T = (2 u sin θ)/g.

       Maximum height is the height of place at which the vertical component of velocity becomes zero or it is the vertical displacement during time of ascent.

      Substituting t = t_a = (u sin θ)/g and y = H_max in y = (u sin θ) t -   gt²

The formula for maximum height H_max = (u² sin² θ)/2g.

Horizontal range is the maximum horizontal displacement of a projectile when it reaches the same horizontal level of projection i.e, it is the horizontal displacement during the time of flight. So, t = T = (2u sinθ)/g when x = horizontal range R. Substituting in

                      X = (u cosθ)t

                      R = (u² sin 2θ)/g

6. Show that the path of a horizontal projectile is a parabola.

A: When an object is thrown horizontally with an initial velocity u from a height h its horizontal displacement s = x in time t, a = 0 along horizontal direction neglecting air resistance. Substituting in

s = ut +  

 at² 

 x = ut.

Its vertical initial velocity u = 0,

Vertical acceleration a = g

Vertical displacement s = y in time t = t.

Substituting in      

 s = ut +    at²

Which represents a parabola.

7. What is the final velocity of an oblique projectile after time t of projection?

A: The horizontal component of velocity of the projectile is u cosθ which remains constant. But its vertical component changes. Vertical initial velocity u_y = u sinθ,   a_y = -g.

Substituting in V_y = u_y + a_yt                   We get V_y = u sin θ - gt

8. What is the angle of projection for maximum horizontal range and state the expression for maximum horizontal range?

A: The angle of projection for maximum horizontal range is θ = 45°. Expression for maximum horizontal range is R_max = u²/g.

9. What are the two angles of projection for the same horizontal range?

A: θ and (90°- θ) are the two angles of projection for the same horizontal range of a projectile.