Question

The equation of projectile is $y=16x-\frac{5x^2}{4}$. The horizontal range is:

• A. $16m$
• B. $8m$
• C. $3.2m$
• D. $12.8m$

Answer 1

The correct option is C $3.2m$

let u be the initial speed, a projectile from the origin at an angle $\theta$

eq of projectile is

$y=xtan\theta -\frac{g{x}^2}{2u^2(cos\theta {)}^2}$ eq(1)..

by comparing given eq in question to eq(1)

$tan\theta =16,u^2(cos\theta {)}^2=4$

$sin\theta =\frac{4}{\sqrt{257}},cos\theta =\frac{1}{\sqrt{257}},u=2\sqrt{257}$

we know the range of projectile motion in two dimension is

$R=\frac{2u^{2}sin\theta cos\theta }{g}=\frac{(2\sqrt{257}{)}^2\times \frac{4}{\sqrt{257}}\times \frac{1}{\sqrt{257}}}{10}=\frac{32}{10}=3.2m$

Hence the C option is correct.