9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion.

Question

9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion.

Asked on April 2, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-04-02T14:59:59.440Z

Here are the solutions to the questions: 9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion. The velocity-time graph for a body projected vertically upwards and landing at the base of the building will show: • An initial positive velocity (upwards). • The velocity decreases linearly to zero as the body reaches its maximum height. • The velocity then becomes negative (downwards) and increases linearly in magnitude as the body falls. • The slope of the graph is constant and negative, representing the acceleration due to gravity ($-g$). • Since it lands at the base of the building (below the starting point), the final negative velocity will have a greater magnitude than the initial positive velocity. ` ^ Velocity (v) | | / | / | / |/ +---------------------> Time (t) |\ | \ | \ | \ v ` 10. A student heated equal amount of water in two aluminium containers A and B by a flame of equal hotness. If A was bigger than B, in which container will it take longer time to boil the water and why? Container A will take longer to boil the water. This is because container A is bigger, meaning it has a larger mass of aluminium. Since both containers are made of the same material, container A will have a larger heat capacity than container B. Therefore, more heat energy will be absorbed by container A itself to reach the boiling temperature of water, even though the mass of water is the same in both. As the flame provides heat at the same rate, the system with the larger total heat capacity (container A + water) will require more time to reach the boiling point. 11. 0.2 kg of copper at $80^\circ\text{C}$ is put in a well lagged brass calorimeter of mass 0.1 kg containing 0.16 kg of sea water at $20^\circ\text{C}$. Calculate the final steady temperature of the mixture. Take specific heat capacity of Copper = $400 \text{ J kg}^{-1} \text{ K}^{-1}$ Brass = $380 \text{ J kg}^{-1} \text{ K}^{-1}$ Sea water = $3900 \text{ J kg}^{-1} \text{ K}^{-1}$ According to the principle of calorimetry, heat lost by the hot body equals heat gained by the cold bodies. Let $T_f$ be the final steady temperature. Heat lost by copper = $m_c c_c (T_c - T_f)$ Heat gained by brass calorimeter = $m_b c_b (T_f - T_b)$ Heat gained by sea water = $m_w c_w (T_f - T_w)$ Given values: $m_c = 0.2 \text{ kg}$, $T_c = 80^\circ\text{C}$, $c_c = 400 \text{ J kg}^{-1} \text{ K}^{-1}$ $m_b = 0.1 \text{ kg}$, $T_b = 20^\circ\text{C}$, $c_b = 380 \text{ J kg}^{-1} \text{ K}^{-1}$ $m_w = 0.16 \text{ kg}$, $T_w = 20^\circ\text{C}$, $c_w = 3900 \text{ J kg}^{-1} \text{ K}^{-1}$ Step 1: Set up the energy balance equation. $$m_c c_c (T_c - T_f) = m_b c_b (T_f - T_b) + m_w c_w (T_f - T_w)$$ Step 2: Substitute the given values into the equation. $$(0.2)(400)(80 - T_f) = (0.1)(380)(T_f - 20) + (0.16)(3900)(T_f - 20)$$ Step 3: Simplify the equation. $$80(80 - T_f) = 38(T_f - 20) + 624(T_f - 20)$$ $$6400 - 80T_f = (38 + 624)(T_f - 20)$$ $$6400 - 80T_f = 662(T_f - 20)$$ $$6400 - 80T_f = 662T_f - 13240$$ Step 4: Rearrange the equation to solve for $T_f$. $$6400 + 13240 = 662T_f + 80T_f$$ $$19640 = 742T_f$$ $$T_f = \frac{19640}{742}$$ $$T_f \approx 26.46899^\circ\text{C}$$ Step 5: Round to an appropriate number of significant figures. $$T_f \approx 26.5^\circ\text{C}$$ The final steady temperature of the mixture is $\boxed{\text{26.5}^\circ\text{C}}$. 12. State two features that make the clinical thermometer more sensitive. Two features that make a clinical thermometer more sensitive are: • A narrow bore (capillary tube) ensures that a small change in the volume of mercury results in a significant change in the length of the mercury column, making it easier to read small temperature variations. • A thin glass bulb allows for faster heat transfer from the body to the mercury, enabling the thermometer to respond quickly to temperature changes. 13. (a) The figure below represents a tube through which a liquid is flowing as shown by the arrow. On the diagram show the relative positions of the level of the liquid in sections marked X, Y and Z. According to Bernoulli's principle, for a horizontal flow of an incompressible fluid, where the velocity of the fluid is higher, the pressure is lower. • In section X, the tube is wide, so the liquid velocity is relatively low, and thus the pressure is high. • In section Y, the tube is narrowest, so the liquid velocity is highest, and thus the pressure is lowest. • In section Z, the tube is wider than Y but narrower than X, so the liquid velocity is intermediate, and the pressure is also intermediate (higher than Y, lower than X). Therefore, the liquid level in the vertical tubes will be highest at X, lowest at Y, and intermediate at Z. ` ^ | | X | | | | | | | | Z | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |

Accepted Answer · 2026-04-02T14:59:59.440Z

Here are the solutions to the questions: 9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion. The velocity-time graph for a body projected vertically upwards and landing at the base of the building will show: • An initial positive velocity (upwards). • The velocity decreases linearly to zero as the body reaches its maximum height. • The velocity then becomes negative (downwards) and increases linearly in magnitude as the body falls. • The slope of the graph is constant and negative, representing the acceleration due to gravity ($-g$). • Since it lands at the base of the building (below the starting point), the final negative velocity will have a greater magnitude than the initial positive velocity. ` ^ Velocity (v) | | / | / | / |/ +---------------------> Time (t) |\ | \ | \ | \ v ` 10. A student heated equal amount of water in two aluminium containers A and B by a flame of equal hotness. If A was bigger than B, in which container will it take longer time to boil the water and why? Container A will take longer to boil the water. This is because container A is bigger, meaning it has a larger mass of aluminium. Since both containers are made of the same material, container A will have a larger heat capacity than container B. Therefore, more heat energy will be absorbed by container A itself to reach the boiling temperature of water, even though the mass of water is the same in both. As the flame provides heat at the same rate, the system with the larger total heat capacity (container A + water) will require more time to reach the boiling point. 11. 0.2 kg of copper at $80^\circ\text{C}$ is put in a well lagged brass calorimeter of mass 0.1 kg containing 0.16 kg of sea water at $20^\circ\text{C}$. Calculate the final steady temperature of the mixture. Take specific heat capacity of Copper = $400 \text{ J kg}^{-1} \text{ K}^{-1}$ Brass = $380 \text{ J kg}^{-1} \text{ K}^{-1}$ Sea water = $3900 \text{ J kg}^{-1} \text{ K}^{-1}$ According to the principle of calorimetry, heat lost by the hot body equals heat gained by the cold bodies. Let $T_f$ be the final steady temperature. Heat lost by copper = $m_c c_c (T_c - T_f)$ Heat gained by brass calorimeter = $m_b c_b (T_f - T_b)$ Heat gained by sea water = $m_w c_w (T_f - T_w)$ Given values: $m_c = 0.2 \text{ kg}$, $T_c = 80^\circ\text{C}$, $c_c = 400 \text{ J kg}^{-1} \text{ K}^{-1}$ $m_b = 0.1 \text{ kg}$, $T_b = 20^\circ\text{C}$, $c_b = 380 \text{ J kg}^{-1} \text{ K}^{-1}$ $m_w = 0.16 \text{ kg}$, $T_w = 20^\circ\text{C}$, $c_w = 3900 \text{ J kg}^{-1} \text{ K}^{-1}$ Step 1: Set up the energy balance equation. $$m_c c_c (T_c - T_f) = m_b c_b (T_f - T_b) + m_w c_w (T_f - T_w)$$ Step 2: Substitute the given values into the equation. $$(0.2)(400)(80 - T_f) = (0.1)(380)(T_f - 20) + (0.16)(3900)(T_f - 20)$$ Step 3: Simplify the equation. $$80(80 - T_f) = 38(T_f - 20) + 624(T_f - 20)$$ $$6400 - 80T_f = (38 + 624)(T_f - 20)$$ $$6400 - 80T_f = 662(T_f - 20)$$ $$6400 - 80T_f = 662T_f - 13240$$ Step 4: Rearrange the equation to solve for $T_f$. $$6400 + 13240 = 662T_f + 80T_f$$ $$19640 = 742T_f$$ $$T_f = \frac{19640}{742}$$ $$T_f \approx 26.46899^\circ\text{C}$$ Step 5: Round to an appropriate number of significant figures. $$T_f \approx 26.5^\circ\text{C}$$ The final steady temperature of the mixture is $\boxed{\text{26.5}^\circ\text{C}}$. 12. State two features that make the clinical thermometer more sensitive. Two features that make a clinical thermometer more sensitive are: • A narrow bore (capillary tube) ensures that a small change in the volume of mercury results in a significant change in the length of the mercury column, making it easier to read small temperature variations. • A thin glass bulb allows for faster heat transfer from the body to the mercury, enabling the thermometer to respond quickly to temperature changes. 13. (a) The figure below represents a tube through which a liquid is flowing as shown by the arrow. On the diagram show the relative positions of the level of the liquid in sections marked X, Y and Z. According to Bernoulli's principle, for a horizontal flow of an incompressible fluid, where the velocity of the fluid is higher, the pressure is lower. • In section X, the tube is wide, so the liquid velocity is relatively low, and thus the pressure is high. • In section Y, the tube is narrowest, so the liquid velocity is highest, and thus the pressure is lowest. • In section Z, the tube is wider than Y but narrower than X, so the liquid velocity is intermediate, and the pressure is also intermediate (higher than Y, lower than X). Therefore, the liquid level in the vertical tubes will be highest at X, lowest at Y, and intermediate at Z. ` ^ | | X | | | | | | | | Z | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |

9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion.

9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion.

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9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion.

9. A body is projected vertically upwards from the top of a building. Assuming that it lands at the base of the building. Sketch the velocity-time graph for this motion.

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