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Q3.26:
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.
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.
Solution :
Step 1: Establishing Kinematic Equations
Let the top of the cliff be the origin ($y = 0$) and the upward direction be positive. The position $y(t)$ of an object under gravity is given by: $y(t) = u_i t - \frac{1}{2}gt^2$ [Per the Second Equation of Motion].
For Stone 1 ($u_1 = 15 \text{ m/s}$): $y_1(t) = 15t - 5t^2$
For Stone 2 ($u_2 = 30 \text{ m/s}$): $y_2(t) = 30t - 5t^2$
Step 2: Defining Relative Position ($x_{rel}$)
The relative position of the second stone with respect to the first is defined as: $x_{rel} = y_2(t) - y_1(t)$
$x_{rel} = (30t - 5t^2) - (15t - 5t^2) = 15t$
This linear relationship holds as long as both stones are in the air [Algebraic simplification].
Step 3: Calculating Time of Flight ($T$)
Stones hit the ground at $y = -200 \text{ m}$. For Stone 1: $-200 = 15t_1 - 5t_1^2 \implies t_1^2 - 3t_1 - 40 = 0$. Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t_1 = \frac{3 + \sqrt{9 + 160}}{2} \approx 7.9 \text{ s}$ (Taking positive root).
For Stone 2: $-200 = 30t_2 - 5t_2^2 \implies t_2^2 - 6t_2 - 40 = 0$. $t_2 = \frac{6 + \sqrt{36 + 160}}{2} \approx 10.2 \text{ s}$.
Step 4: Analyzing the Motion Phases
| Time Interval | Relative Position ($x_{rel}$) | Justification |
|---|---|---|
| $0 \leq t \leq 7.9 \text{ s}$ | $15t$ | Both stones in flight. |
| $7.9 < t \leq 10.2 \text{ s}$ | $y_2(t) - (-200) = 30t - 5t^2 + 200$ | Stone 1 stationary at base. |
| $t > 10.2 \text{ s}$ | $0$ | Both stones stationary at base. |
Step 5: Visual Representation
Final Analysis
The graph consists of two distinct segments:
1. Linear Segment ($0 \leq t \leq 7.9 \text{ s}$): $x_{rel} = 15t$. This is a straight line passing through the origin with a slope of $15 \text{ m/s}$.
2. Parabolic Segment ($7.9 < t \leq 10.2 \text{ s}$): $x_{rel} = -5t^2 + 30t + 200$. This represents the position of the second stone relative to the grounded first stone.
Final Solution: The relative position is defined by $x_{rel} = 15t$ for $0 \leq t \leq 7.9$ s and $x_{rel} = -5t^2 + 30t + 200$ for $7.9 < t \leq 10.2$ s.
More Questions from Class 11 Physics Motion in a Straight Line Additional Exercises
- Q3.23: 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 ?
- Q3.24: 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 ?
- Q3.25(a): On a long horizontally moving belt (Fig. 3.26), 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 ?.
- Q3.25(b): On a long horizontally moving belt (Fig. 3.26), 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 (b) speed of the child running opposite to the direction of motion of the belt ?
- Q3.25(c): On a long horizontally moving belt (Fig. 3.26), 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 (c) time taken by the child in (a) speed of the child running in the direction of motion of the belt ? and (b) speed of the child running opposite to the direction of motion of the belt ? Which of the answers alter if motion is viewed by one of the parents ?
- Q3.27(a): The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) $t = 0 s$ to $10 s$.
- Q3.27(b): The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (b) $t = 2 s$ to $6 s$.
- Q3.27(c): The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. What is the average speed of the particle over the interval in (a) $t = 0 s$ to $10 s$?
- Q3.27(d): The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. What is the average speed of the particle over the interval in (b) $t = 2 s$ to $6 s$?
- Q3.28(a): The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 : Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$: (a) $x(t_2) = x(t_1) + v(t_1) (t_2 - t_1) +(\frac{1}{2}) a (t_2 - t_1)^2$
- Q3.28(b): The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 : Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$: (b) $v(t_2) = v(t_1) + a (t_2 - t_1)$
- Q3.28(c): The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 : Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$: (c) $v_{average} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$
- Q3.28(d): The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 : Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$: (d) $a_{average} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$
- Q3.28(e): The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 : Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$: (e) $x(t_2) = x(t_1) + v_{average} (t_2 - t_1) + (\frac{1}{2}) a_{average} (t_2 - t_1)^2$
- Q3.28(f): The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 : Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$: (f) $x(t_2) - x(t_1)$ = area under the $v-t$ curve bounded by the $t$-axis and the dotted line shown.
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