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