Q145 / 160 · TS EAMCET 2017 · Physics

A ball is thrown at a speed of $20 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ with the horizontal. The maximum height reached by the ball is (Use $g=10 \mathrm{~m} / \mathrm{s}^2$ )

  • A
    $2 \mathrm{~m}$
  • B
    $3 \mathrm{~m}$
  • C
    $4 \mathrm{~m}$
  • Correct answer
    D
    $5 \mathrm{~m}$

Solution

Given, Speed of ball $(u)=20 \mathrm{~m} / \mathrm{s}$ Angle $(\theta)=30^{\circ}$ We know that, $$ \begin{aligned} & H=\frac{u^2 \sin ^2 \theta}{2 g} \\ & H=\frac{(20)^2(\sin 30)^2}{2 \times 10} \\ & \Rightarrow \quad H=\frac{20 \times 20 \times\left(\frac{1}{2}\right)^2}{20} \\ & H=20 \times \frac{1}{4} \\ & \Rightarrow \quad H=5 \mathrm{~m} \\ & \end{aligned} $$

Hence, the correct answer is option: 'D'.

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