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Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane3 Marks
Question
A projectile is projected with velocity $u$ making an angle $\theta$ with horizontal direction, find:
  1. Time of flight.
  2. Horizontal range.
✓

Answer

  1. At maximum height $v =0$
$\therefore\ 0=\text{u}\sin\theta-\text{gt};$
or $\text{t}=\frac{\text{u}\sin\theta}{\text{g}}$
But time of flight,
$\text{T}=\text{2t}=\frac{2\text{u}\sin\theta}{\text{g}}$
  1. When the body returns to the same horizontal level $y = 0$
$\therefore 0=\text{x}\tan\theta-\frac{\text{gx}^2}{2\text{u}^2\cos^2\theta} [$From $(i)]$
or $\text{x}\tan\theta=\frac{\text{gx}^2}{2\text{u}^2\cos^2\theta}$
or $\text{x}=\frac{2\text{u}^2\sin\theta\cos\theta}{\text{g}}=\frac{\text{u}^2\sin2\theta}{\text{g}}$
But coordinates of $M$ are $(R, 0).$
Putting $x = R$, we have $\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}$

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