Ncert Exercises & Solutions

4.15 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall?

NEETprep Answer:

W e k n o w t h a t,

$W e k n o w t h a t,$

H = \frac{u^{2} \sin^{2} θ}{2 g}

$H = \frac{u^{2} \sin^{2} θ}{2 g}$

\Rightarrow 25 = \frac{{(40)}^{2} \sin^{2} θ}{2 \times 9.8}

$\Rightarrow 25 = \frac{{(40)}^{2} \sin^{2} θ}{2 \times 9.8}$

\Rightarrow \sin θ = \sqrt{\frac{25 \times 2 \times 9.8}{{(40)}^{2}}} = 0.55334

$\Rightarrow \sin θ = \sqrt{\frac{25 \times 2 \times 9.8}{{(40)}^{2}}} = 0.55334$

\Rightarrow θ = 33 . 6^{0}

$\Rightarrow θ = 33 . 6^{0}$

R a n g e, R = \frac{u^{2} \sin 2 θ}{g} = \frac{{(40)}^{2} \times \sin (2 \times 33 . 6^{0})}{9.8} = 150.53 m

$R a n g e, R = \frac{u^{2} \sin 2 θ}{g} = \frac{{(40)}^{2} \times \sin (2 \times 33 . 6^{0})}{9.8} = 150.53 m$

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