Question

Trajectory of particle in a projectile is given as y = x – x 2 80 . Here, x and y are in metre. For this projectile motion match the following and select proper option (g = 10 m/s2 ) Column-I Column-II A. Angle of projection p. 20 m B. Angle of velocity with q. 80 m horizontal after 4s C. Maximum height r. 45° D. Horizontal range s. tan–1 1 2 F H G I K

Answer 1

Given:

Trajectory of particle is y = x - 80x².

To Match the columns:

Solution:

This Equation belongs to a projectile trajectory;

$\boxed{y = x - 80 {x}^{2} }$

Putting y = 0 , for range calculation:

$\therefore \: 0 = x - 80 {x}^{2}$

$= > \: 80 {x}^{2} = x$

$= > \: 80 x = 1$

$\boxed{ = > \:x = \dfrac{1}{80} \: m = range}$

For maximum height, putting x = range/2:

$y = x - 80 {x}^{2}$

$= > y = \dfrac{1}{160} - 80 { (\dfrac{1}{160} )}^{2}$

$= > y = \dfrac{1}{160} \{1 - 80 (\dfrac{1}{160} )\}$

$= > y = \dfrac{1}{160} \{1 - (\dfrac{1}{2} )\}$

$\boxed{= > y = \dfrac{1}{320} \: m = max \: height}$

For angle of Projection:

$\tan( \theta) = 1$

$\boxed{= > \theta = 45 \degree }$

Hope It Helps.