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 sn = 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 = gta. Time of ascent ta = u/g. If the maximum height reached is hmax i.e., y = hmax substituting in
y = ut - gt2 hmax = uta - gta2 = (gta) ta - gta2 = gta2
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 (td). From y = gt2
hmax = gtd2 =
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 = ta = (u sin θ)/g and y = Hmax in y = (u sin θ) t - gt2
The formula for maximum height Hmax = (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 uy = u sinθ, ay = -g.
Substituting in Vy = uy + ayt We get Vy = 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 Rmax = 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.