Question

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s¯¹ and 30 m s¯¹. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s¯². Give the equations for the linear and curved parts of the plot.

Answer 1

Given: The height of the cliff is 200 m, the initial speed of first stone is 15 m/s, the initial speed of second stone is 30 m/s and the acceleration due to gravity is 10 m/ s 2 .

The stones are thrown upwards and move against the gravity of Earth, therefore, the acceleration acting on the stone will be a=−g=−10 m/ s 2

Consider the first stone,

The vertical height attained by first stone is given as,

x 1 = x 0 + u 1 t+ 1 2 a t 2

Where, x 1 is the height attained by first stone, x 0 is the height of the cliff, u 1 is the initial velocity of stone 1, t is the taken by stone 1 and a is the acceleration of the stone 1.

By substituting the values in the above expression, we get

x 1 =200+15t+ 1 2 ( −10 ) t 2 x 1 =200+15t−5 t 2 (I)

When the first stone hits the ground, x 1 =0

By substituting the values in the above expression, we get

200+15t−5 t 2 =0 t 2 −3t−40=0 t( t−8 )+5( t−8 )=0 t=8 or −5

The time cannot be negative. Hence, t=8 s.

Consider the second stone,

The vertical height attained by second stone is given as,

x 2 = x 0 + u 2 t+ 1 2 a t 2

Where, x 2 is the height attained by second stone, x 0 is the height of the cliff, u 2 is the initial velocity of stone 2, t is the taken by stone 2 and a is the acceleration of the stone 2.

By substituting the values in the above expression, we get

x 2 =200+30t+ 1 2 ( −10 ) t 2 x 2 =200+30t−5 t 2 (II)

When the second stone hits the ground, x 2 =0

By substituting the values in the above expression, we get

200+30t−5 t 2 =0 t 2 −6t−40=0 t( t−10 )+4( t−10 )=0 t=10 or −4

The time cannot be negative. Hence, t=10 s.

Subtract equation (I) from equation (II),

x 2 − x 1 =( 200+30t−5 t 2 )−( 200+15t−5 t 2 ) x 2 − x 1 =15t (III)

It is clear from equation (III) that, the equation (III) represents the linear path of both stones. Hence, the path remains a straight line till t=8 s.

The maximum separation between stones at t=8 s is given as,

x 2 − x 1 =15t

By substituting the values in the above expression, we get

x 2 − x 1 =15×8 =120

It is clear from the above calculation for separation between two stones that the graph correctly represents the time variation of the relative position of the second stone with respect to the first.

After t=8 s, only second stone is in motion whose variation with time is represented by the following quadratic equation.

x 2 − x 1 =( 200+30t−5 t 2 )

Hence, the equation of linear is ( x 2 − x 1 )=15t and the equation of curved path is ( x 2 − x 1 )=( 200+30t−5 t 2 ).