Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of \( 15ms^{-1}\) and \(30ms^{-1}\)Verify that the graph shown in given figure 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 = \( 10ms^{-2}\) Give the equations for the linear and curved parts of the plot.

Asked by Abhisek | 1 year ago | 290

For the first stone:

Given,

Acceleration, a = –g = – 10 m/s^{2}

Initial velocity, u_{I} = 15 m/s

Now, we know

s_{1} = s_{0} + u_{1}t + (\( \dfrac{1}{2}\))at^{2}

Given, height of the tree, s_{0} = 200 m

s_{1} = 200 + 15t – 5t^{2 }. . . . . . . . . . ( 1 )

When this stone hits the jungle floor, s_{1} = 0

– 5t^{2 }+ 15t + 200 = 0

t^{2} – 3t – 40 = 0

t^{2} – 8t + 5t – 40 = 0

t (t – 8) + 5 (t – 8) = 0

t = 8 s or t = – 5 s

Since, the stone was thrown at time t = 0, the negative sign is not possible

t = 8 s

For second stone:

Given,

Acceleration, a = – g = – 10 m/s^{2}

Initial velocity, u_{II }= 30 m/s

We know,

s_{2} = s_{0} + u_{II}t + (\( \dfrac{1}{2}\))at^{2}

= 200 + 30t – 5t^{2} . . . . . . . . . . . . . ( 2 )

when this stone hits the jungle floor; s_{2} = 0

– 5t^{2} + 30 t + 200 = 0

t^{2} – 6t – 40 = 0

t^{2} – 10t + 4t + 40 = 0

t (t – 10) + 4 (t – 10) = 0

(t – 10) (t + 4) = 0

t = 10 s or t = – 4 s

Here again, the negative sign is not possible

t = 10 s

Subtracting equations ( 1 ) from equation ( 2 ), we get

s_{2} – s_{1} = (200 + 30t -5t^{2}) – (200 + 15t -5t^{2})

s_{2} – s_{1} =15t . . . . . . . . . . . . .. . . . . . ( 3 )

Equation ( 3 ) represents the linear trajectory of the two stone, because to this

linear relation between (s_{2 }– s_{1}) and t,, the projection is a straight line till 8 s.

The maximum distance between the two stones is at t = 8 s.

(s_{2} – s_{1})_{max }= 15× 8 = 120 m

This value has been depicted correctly in the above graph.

After 8 s, only the second stone is in motion whose variation with time is given by the quadratic equation:

s_{2} – s_{1 }= 200 + 30t – 5t^{2}

Therefore, the equation of linear and curved path is given by :

s_{2 }– s_{1} = 15t (Linear path)

s_{2} – s_{1} = 200 + 30t – 5t^{2} (Curved path)

The velocity-time graph of a particle in one-dimensional motion is shown in the figure.

**(a)** \( x (t_2) = x (t_1) + v (t_1) (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a(t_2 – t_1)^2\)

**(b)** \( v(t_2) = v(t_1) + a(t_2 – t_1)\)

**(c)** Vaverage = \( \dfrac{ x(t_2) – x (t_1)}{(t_2 – t_1)}\)

**(d)** aaverage = \( \dfrac{v(t_2) – v (t_1) }{(t_2 – t_1)}\)

**(e)** \( x (t_2) = x (t_1) + v_{av} (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a_{av} (t_2 – t_1)^2\)

**(f)** \( x(t_2) – x (t_1)\) = Area under the v-t curve bounded by t- axis and the dotted lines.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

**(a)** t = 0 s to 10 s,

**(b)** t = 2 s to 6 s.

On a long horizontally moving belt figure, a child runs to and fro with a speed 9 km h^{–1} (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^{–1}. For an observer on a stationary platform outside, what is the

**(a)** speed of the child running in the direction of motion of the belt ?.

**(b)** speed of the child running opposite to the direction of motion of the belt?

**(c)** time taken by the child in (a) and (b)?

Which of the answers alter if motion is viewed by one of the parents?

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^{-1}. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s^{-1} and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?

A three-wheeler starts from rest, accelerates uniformly with 1 m s^{–2} on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?