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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line1 Mark
MCQ
The object is released from rest under gravity at $y = 0$. The equation of motion which correctly expresses the above situation is:
  • A
    $\text{v} = -9.8\text{t ms}^{-1}$
  • B
    $\text{v} = (9.8 - 9.8\text{t)m/s}$
  • ✓
    $\text{v}^2 = -19.6\text{y}^2 \text{m}^2\text{s}^{-2}$
  • D
    $\text{v}^2=(\text{v}^2_0+29.6\text{y})\text{m}^2/\text{s}^2$
✓

Answer

Correct option: C.
$\text{v}^2 = -19.6\text{y}^2 \text{m}^2\text{s}^{-2}$
For free fall, $v_0 = 0, a = -g = -9.8\ ms^{-1}$
The equations of motion are
$v = -9.8t\ ms^{-1} ($using $v = v_0 + at)$
$v^2 = 2(-9.8)y ($using $\text{v}^2=\text{v}^2_0+2\text{ay})$
$= -19.6y\ m^2s^{-2}$

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