Ncert Exercises & Solutions

4.16 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

NEETprep Answer:

$F o r m a x i m u m r a n g e, θ = 45^{0}$

$S i n c e, R = \frac{u^{2} \sin 2 θ}{g}$

$\Rightarrow R_{m a x} = \frac{u^{2} \sin 90^{0}}{g} = \frac{u^{2}}{g}$

$\Rightarrow 100 = \frac{u^{2}}{g} . . . . . (1)$

$H = \frac{u^{2} \sin^{2} θ}{2 g}$

$\Rightarrow H_{m a x} = \frac{u^{2} \sin^{2} 90^{0}}{2 g}$

$\Rightarrow H_{m a x} = \frac{u^{2}}{2 g} . . . . . (2)$

$F r o m (1) a n d (2),$

$H_{m a x} = \frac{100}{2} = 50 m$