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Q1 [3 Marks]: Draw the position-time ($x-t$) and velocity-time ($v-t$) graphs for an object dropped from a height and falling freely under gravity.
Part A: Position-Time ($x-t$) Graph
Let the initial height be $H$ and the ground be at $x = 0$. Since the object is dropped ($u=0$) and accelerates downwards under gravity ($a = -g$), its position is given by $x(t) = H - \frac{1}{2}gt^2$.
This represents a downward-opening parabola starting from $x=H$ at $t=0$, curving downwards until it hits the t-axis ($x=0$).
Let the initial height be $H$ and the ground be at $x = 0$. Since the object is dropped ($u=0$) and accelerates downwards under gravity ($a = -g$), its position is given by $x(t) = H - \frac{1}{2}gt^2$.
This represents a downward-opening parabola starting from $x=H$ at $t=0$, curving downwards until it hits the t-axis ($x=0$).
Part B: Velocity-Time ($v-t$) Graph
The velocity of the falling object at any time $t$ is given by $v = u + at$. Since $u=0$ and $a=-g$, the equation is $v(t) = -gt$.
This is a straight line passing through the origin with a constant negative slope (slope $= -g$). It moves into the negative quadrant indicating downward direction.
The velocity of the falling object at any time $t$ is given by $v = u + at$. Since $u=0$ and $a=-g$, the equation is $v(t) = -gt$.
This is a straight line passing through the origin with a constant negative slope (slope $= -g$). It moves into the negative quadrant indicating downward direction.
Position-Time (x-t) Graph
Velocity-Time (v-t) Graph
Q2 [3 Marks]: A body falling freely under gravity passes two points A and B separated by $40\text{ m}$ in $2\text{ s}$. Find the height of point A above the starting point.
Step 1: Analyze the motion from A to B.
Let the velocity of the body when it reaches point A be $v_A$. The distance between A and B is $s = 40\text{ m}$, and the time taken is $t = 2\text{ s}$. Taking $g = 10\text{ m/s}^2$ downward as positive.
Let the velocity of the body when it reaches point A be $v_A$. The distance between A and B is $s = 40\text{ m}$, and the time taken is $t = 2\text{ s}$. Taking $g = 10\text{ m/s}^2$ downward as positive.
$$s = v_A t + \frac{1}{2}gt^2$$
$$40 = v_A(2) + \frac{1}{2}(10)(2)^2$$
$$40 = 2v_A + 20$$
$$20 = 2v_A \Rightarrow v_A = 10\text{ m/s}$$
Step 2: Analyze the motion from the starting point to A.
The body is dropped from rest, so initial velocity $u = 0$. The final velocity at point A is $v_A = 10\text{ m/s}$. Let the height of A above the start be $H_A$.
The body is dropped from rest, so initial velocity $u = 0$. The final velocity at point A is $v_A = 10\text{ m/s}$. Let the height of A above the start be $H_A$.
$$v_A^2 = u^2 + 2gH_A$$
$$(10)^2 = 0^2 + 2(10)H_A$$
$$100 = 20H_A \Rightarrow H_A = \frac{100}{20}$$
$$\text{Height of point A from start} = 5\text{ m}$$
Q3 [4 Marks]: A juggler throws balls into the air. He throws one whenever the previous one is at its highest point. If he throws $n$ balls each second, find the maximum height reached by each ball.
Step 1: Determine the time interval between throws.
Since the juggler throws $n$ balls per second, the time gap between two consecutive throws is:
Since the juggler throws $n$ balls per second, the time gap between two consecutive throws is:
$$\Delta t = \frac{1}{n} \text{ seconds}$$
Step 2: Relate time interval to the ball's flight.
The problem states he throws a new ball when the previous one is at its maximum height. Therefore, the time taken for a ball to reach its maximum height ($t_{up}$) is exactly equal to the time gap between throws.
The problem states he throws a new ball when the previous one is at its maximum height. Therefore, the time taken for a ball to reach its maximum height ($t_{up}$) is exactly equal to the time gap between throws.
$$t_{up} = \frac{1}{n} \text{ seconds}$$
Step 3: Find the initial velocity ($u$).
At maximum height, final velocity $v = 0$. Using $v = u - gt_{up}$:
At maximum height, final velocity $v = 0$. Using $v = u - gt_{up}$:
$$0 = u - g\left(\frac{1}{n}\right) \Rightarrow u = \frac{g}{n}$$
Step 4: Calculate the maximum height ($H$).
Use the formula for maximum height $H = \frac{u^2}{2g}$.
Use the formula for maximum height $H = \frac{u^2}{2g}$.
$$H = \frac{\left(\frac{g}{n}\right)^2}{2g} = \frac{\frac{g^2}{n^2}}{2g}$$
$$H = \frac{g^2}{2g n^2}$$
$$\text{Maximum Height } (H) = \frac{g}{2n^2}$$
Q4 [5 Marks]: A stone is dropped from the top of a cliff. It is seen to pass a window $2.45\text{ m}$ high in $0.2\text{ s}$. How far is the top of the window below the top of the cliff? ($g = 9.8\text{ m/s}^2$).
Step 1: Analyze motion past the window.
Let the velocity of the stone as it crosses the top of the window be $u_w$.
Distance $s = 2.45\text{ m}$, Time $t = 0.2\text{ s}$, $a = 9.8\text{ m/s}^2$.
Let the velocity of the stone as it crosses the top of the window be $u_w$.
Distance $s = 2.45\text{ m}$, Time $t = 0.2\text{ s}$, $a = 9.8\text{ m/s}^2$.
$$s = u_w t + \frac{1}{2}gt^2$$
$$2.45 = u_w(0.2) + \frac{1}{2}(9.8)(0.2)^2$$
$$2.45 = 0.2u_w + 4.9(0.04)$$
$$2.45 = 0.2u_w + 0.196$$
Step 2: Solve for $u_w$.
$$0.2u_w = 2.45 - 0.196 = 2.254$$
$$u_w = \frac{2.254}{0.2} = 11.27\text{ m/s}$$
This is the velocity of the stone right at the top edge of the window.
Step 3: Analyze motion from the cliff top to the top of the window.
The stone was dropped from the cliff, so initial velocity $u_{cliff} = 0$.
Final velocity at the window $v = u_w = 11.27\text{ m/s}$.
Let the distance from the cliff top to the window top be $H$.
The stone was dropped from the cliff, so initial velocity $u_{cliff} = 0$.
Final velocity at the window $v = u_w = 11.27\text{ m/s}$.
Let the distance from the cliff top to the window top be $H$.
$$v^2 = u^2 + 2gH$$
$$(11.27)^2 = 0^2 + 2(9.8)H$$
$$127.0129 = 19.6H$$
$$H = \frac{127.0129}{19.6}$$
$$\text{Distance to top of window } (H) = 6.48\text{ m}$$
