Ncert Exercises & Solutions

The horizontal range of a projectile fired at an angle of $15^\circ$ is $50$ m. If it is fired with the same speed at an angle of $45^\circ$, its range will be:
1. $60$ m
2. $71$ m
3. $100$ m
4. $141$ m

Step 1: Find the initial speed of the projectile.

We know that
where $θ$ is the angle of projection
Given

$\begin{array}{l} θ = 15^{\circ} and R = 50 m \\ Range, R = \frac{u^{2} \sin 2 θ}{g} \end{array}$

Putting all the given values in the formula, we get

$\begin{array}{r} \Rightarrow A = 50 m = \frac{u^{2} \sin (2 \times 15^{\circ})}{g} \\ \Rightarrow 50 \times g = u^{2} \sin 30^{\circ} = u^{2} \times \frac{1}{2} \end{array}$

$\begin{aligned} \Rightarrow & 50 \times g \times 2 = u^{2} \\ \Rightarrow & u^{2} = 50 \times 9 .8 \times 2 = 100 \times 9 .8 = 980 \\ \Rightarrow & u = \sqrt{980} = \sqrt{49 \times 20} = 7 \times 2 \times \sqrt{5} m / s \\ = 14 \times 223 m / s = 31 .304 m / s \end{aligned}$
$\begin{array}{lll} For & θ = 45^{\circ}; R = \frac{u^{2} \sin 2 \times 45^{\circ}}{g} = \frac{u^{2}}{g} & (∵ \sin 90^{\circ} = 1) \\ \Rightarrow & A = \frac{(14 \sqrt{5})^{2}}{g} = \frac{14 \times 14 \times 5}{9 .8} = 100 m \end{array}$

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