Question Bank • QB_FM_ADV_035 NAT • 2 Marks

Section 2: Farm Machinery

Farm Machinery

Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing

Question Statement:

A seed planter travels forward at a constant speed of $v_f = 7.2\text{ km/h}$. A seed is released from the metering mechanism and drops freely through a vertical seed tube of height $h = 0.45\text{ m}$ into the open furrow. Assuming zero air drag, zero initial horizontal velocity relative to the planter ($v_{x0} = v_f$), and $g = 9.81\text{ m/s}^2$, the horizontal distance traveled by the seed relative to the ground during its fall inside the tube in $\text{cm}$ is ________ (round off to 1 decimal place).

Answer Key:

60.6

Step-by-Step Solution & Explanation

``` Seed Fall Kinematics: v_f ---> [Metering Cup] | \ | \ Parabolic Path h = 0.45 m \ | \ v======o (Furrow floor) |<-- x -->| ``` **Step-by-step Calculation:** 1. Forward speed of planter in $\text{m/s}$: $v_f = \frac{7.2 \times 1000}{3600} = 2.0\text{ m/s}$ 2. Free fall time $t$ for vertical drop height $h = 0.45\text{ m}$: $h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.45}{9.81}} = \sqrt{\frac{0.90}{9.81}} = \sqrt{0.09174} = 0.30289\text{ s}$ 3. Horizontal displacement relative to the ground: $x = v_f \times t = 2.0\text{ m/s} \times 0.30289\text{ s} = 0.60579\text{ m} = 60.58\text{ cm} \approx 60.6\text{ cm}$
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