Question

A projectile is thrown from a point in a vertical plane such that its horizontal and vertical velocity component are $9.8$ m/s and $19.6$ m/s respectively. Its horizontal range is :

• A. $4.9$ m
• B. $9.8$ m
• C. $19.6$ m
• D. $39.2$ m

Answer 1

Answer: (D)

$39.2$ m

let horizontal component of velocity be $v_x=vcos\theta$ and vertical be $v_y=vsin\theta$

therefore velocity is $v=\sqrt{v_x^2+v_y^2}=\sqrt{{9.8}^2+{19.6}^2}=9.8\sqrt{5}$

and $\theta =ta{n}^{-1}\frac{v_y}{v_x}=ta{n}^{-1}\frac{19.6}{9.8}=ta{n}^{-1}2$

Range $R=\frac{v^{2}sin2\theta }{g}=\frac{2v^{2}sin\theta cos\theta }{g}=\frac{2(9.8\sqrt{5}{)}^2\frac{2}{\sqrt{5}}\frac{1}{\sqrt{5}}}{g}=39.2m$