Q30 of 37 Page 25

A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R = .


If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig 4.5), show that it could be achieved by raising the gun to a height at least


h = ∆x


(Hint: This problem can be approached in two different ways:


(i) Refer to the diagram: target T is at horizontal distance x = R + ∆x and below point of projection y = – h.


(ii) From point P in the diagram: Projection at speed vo at an angle θ below horizontal with height h and horizontal range ∆x.)


Speed of the shells = vo


Horizontal component of velocity of shells = vx = v0cosθ


Vertical component of velocity of shells = vy = v0sinθ


Maximum range = R =


We know, maximum range is achieved at θ = 45°


Time of flight = t =


Using kinematic equation,


-h = v0sinθt – gt2


Substituting the value of t,


-h = v0sinθ –


Here, θ = 45° , hence sinθ = cosθ =


h = +





∴h = ∆x


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If |A|= 2 and |B| = 4, then match the relations in column I with the angle θ between A and B in column II

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A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g =10 m/s2.

 31

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.)


 32

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.)