Question

A particle is thrown from a point in a horizontal plane such that its horizontal and vertical velocity component are 9.8 metre per seconds and 19.6 metre per second respectively its horizontal range is

Answer 1

horizontal component of velocity=9.8
u cos theta=9.8m/s
vertical component of velocity =19.6
u sin theta=19.6



so ans is 38.4m

Answer 2

Horizontal Range is 39.2 m

SOLUTION:

The particle is being thrown from a horizontal point, thus the particle will follow projectile motion. The velocity “u” of the particle will have two components.

The horizontal component $=\cos \theta$, and the vertical component $u \sin \theta$

According to the question.

$u \cos \theta=9.8 \frac{m}{s}$

$u \sin \theta=19.6 \frac{m}{s}$

To calculate the range exhibited by the projectile we use the equation for range “R” as

$R=\frac{u^{2} \sin 2 \theta}{g}$

From the trigonometric function we have $\sin 2 \theta=2 \sin \theta \cos \theta$

So,

R=\frac{u^{2} 2 \sin \theta \cos \theta}{g}  

This can be further written as

$R=\frac{2 u \sin \theta u \cos \theta}{g}$

Substituting the values we have

$=\frac{2 \times 19.6 \times 9.8}{9.8}$

Range = 39.2 m