A free response practice question on kinematics in two dimensions was given to you in the post dated 11^{th} February 2010. As promised, I give below a model answer along with the question:

A chopper at an air base is rising vertically up with a velocity of 2 ms^{–1}. When it is at an altitude of 50 m, a packet is thrown down with velocity 8 ms^{–1} (with respect to the chopper), making an angle of 60º with the *vertical*. Neglecting air resistance and assuming that *g* = 10 ms^{–2}, answer the following questions

(a) Draw a diagram to show the nature of the path AB followed by the packet from the moment it leaves the chopper to the moment it hits the ground, as seen by an observer on the ground. Is the path straight, circular or parabolic?.

(b) Will the packet move up initially? Justify your answer.

(c) Calculate the time taken by the packet to reach the ground.

(d) Determine the horizontal distance traveled by the packet before hitting the ground.

(e) Determine the velocity (magnitude as well as direction) with which the packet hits the ground.

(a) The path AB of the packet is shown in the following figure. The path is parabolic.(b) The packet will not move up initially. This is due to the fact that the vertical, downward component (8 cos 60º = 4 ms^{–1}) of the velocity with which the packet is thrown down is greater than the upward velocity (2 ms^{–1}) of the chopper.

[The situation is as though the packet is thrown down from a stationary chopper with vertical (downward) velocity component (4 ms^{–1 }–^{ }2 ms^{–1})^{ }= 2 ms^{–1} and horizontal velocity component 8 sin 60º = 4√3 ms^{–1}].

(c) The time *t* taken by the packet to reach the ground can be found by considering its vertical motion. We have

*s = ut* + ½ *at*^{2} where the vertical displacement *s* = 50 m, vertical (downward) velocity component *u* = 8 cos 60º –^{ }2 = 2 ms^{–1} and vertical (downward) acceleration *a* = *g =* 10 ms^{–2}.

Therefore, 50 = 2*t* + ½ ×10*t*^{2}.

Or, 5*t*^{2} + 2*t *– 50 = 0

This gives *t* = [–2 **± **√(4 – 1000)] /10 = 2.968 s.

(d) The horizontal distance *R* traveled by the packet before hitting the ground is the product of the above time and the *horizontal* velocity component *v*_{horizontal} which is equal to 8 sin 60º (= 4√3 ms^{–1}).

Therefore, *R = *2.968×4√3 = 20.56 m.

(e) The velocity with which the packet hits the ground is the resultant of the horizontal and vertical components. The *horizontal *component of velocity remains unchanged (at 8sin 60º = 4√3 ms^{–1}) throughout the motion of the packet since the gravitational force acts vertically and cannot affect the horizontal motion. The *vertical* component of velocity *v*_{vertical}* *goes on increasing during the fall of the packet in accordance with the equation,

*v*_{vertical}* = u*_{vertical}* + at *where *u*_{vertical} is the initial vertical component of velocity, *a* is the vertical acceleration and *t* is the time of flight of the packet.

Here *u*_{vertical} = 8 cos 60º –^{ }2 = 2 ms^{–1}, *a = g = *10 ms^{–2} and *t =* 2.968 s.

Therefore, *v*_{vertical}* =* 2 + 10×2.968 = 31.68 ms^{–1}.

*v = *√[(4√3)^{2} + (31.68)^{2}] = 32.43 ms^{–1}.* *

The direction of this velocity makes an angle *θ* with the *horizontal* (fig.) and is given by

tan* θ = **v*_{vertical}/* v*_{horizontal} = 31.68/(4√3) = 4.572

Therefore *θ = *77.66º

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