Q. 55.0( 2 Votes )
An aeroplane is flying horizontally with a velocity of 600 kmhr–1 and at the height of 1960 m. When it is vertically at a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is
Aeroplane moves horizontally
Hence horizontal velocity = u = 600 kmhr–1
Height = h = 1960 m
Now, horizontal velocity remains constant as acceleration acts vertically.
For horizontal range AB =
Horizontal Velocity × Time of flight
Let the time of flight = T
Now for, Vertical Motion:
s = h; u = 0; a = g; t = T;
Using equations of motion:
Now, u = 600 kmhr–1 =
Rate this question :
A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Fig. 4.6).
(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).
(b) Time of flight.
(c) β at which range will be maximum.
(Hint: This problem can be solved in two different ways:
(i) Point P at which particle hits the plane can be seen as intersection of its trajectory (parabola) and straight line. Remember particle is projected at an angle (α + β) w.r.t. horizontal.
(ii) We can take x-direction along the plane and y-direction perpendicular to the plane. In that case resolve g (acceleration due to gravity) in two different components, gx along the plane and gy perpendicular to the plane. Now the problem can be solved as two independent motions in x and y directions respectively with time as a common parameter.)
Physics - Exemplar
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed vo and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.
(Hint: (i) After rebound, particle still has speed Vo to start.
(ii) Work out angle particle speed has with horizontal after it rebounds.
(iii) Rest is similar to if particle is projected up the incline.)
Physics - Exemplar