Question

A body is projected at angle 45⁰ to horizontal with velocity 20 m/s from ground. If there is an acceleration in horizontal direction of 2 m/s², then calculate horizontal range of this particle-

• A. 40 m
• B. 48 m
• C. 8 m
• D. 20 m

Answer 1

Answer: (B)

48 m

Answer 2

Q Time of flight does not depend on horizontal acceleration

\ T = $\frac { 2 \mathrm { u } \sin \theta } { \mathrm { g } }$ = $\frac { 2 ( 20 ) 1 / \sqrt { 2 } } { 10 }$ = $2 \sqrt { 2 }$ s

Q R = u_xT + $\frac { 1 } { 2 }$ a_xT²

= (20 . cos 45º) ( $2 \sqrt { 2 }$ ) + $\frac { 1 } { 2 }$ (2)( $2 \sqrt { 2 }$ )²

= 40 + 8 = 48 m