A particle is projected towards the north with speed 20 m/s at an angle 45o with horizontal. Ball gets horizontal acceleration of 7.5 m/s2 towards east due to wind. Range of ball (in meter) minus 42 m will be

Question

A particle is projected towards the north with speed 20 m/s at an angle 45o with horizontal. Ball gets horizontal acceleration of 7.5 m/s2 towards east due to wind. Range of ball (in meter) minus 42 m will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Time of flight: T=(2 u sin θ /g)=2√2 s Range (along north) =(u2 sin 2 θ /g)=40 m Range (along east) =(1/2)aT2=30 m ∴ Range =√302 + 402=50 m

Q. A particle is projected towards the north with speed $20 m / s$ at an angle $4 5^{o}$ with horizontal. Ball gets horizontal acceleration of $7.5 m / s^{2}$ towards east due to wind. Range of ball (in meter) minus $42 m$ will be

Time of flight : $T = \frac{2 u s in θ}{g} = 22 s$
Range (along north) $= \frac{u ^{2} s in 2 θ}{g} = 40 m$
Range (along east) $= \frac{1}{2} a T^{2} = 30 m$
$∴$ Range $= 3 0^{2} + 4 0^{2} = 50 m$

Q. A particle is projected towards the north with speed 20m/s at an angle 45o with horizontal. Ball gets horizontal acceleration of 7.5m/s2 towards east due to wind. Range of ball (in meter) minus 42m will be

Time of flight : T=g2usinθ​=22​s Range (along north) =gu2sin2θ​=40m Range (along east) =21​aT2=30m ∴ Range =302+402​=50m

Q. A particle is projected towards the north with speed $20 m / s$ at an angle $4 5^{o}$ with horizontal. Ball gets horizontal acceleration of $7.5 m / s^{2}$ towards east due to wind. Range of ball (in meter) minus $42 m$ will be

Time of flight : $T = \frac{2 u s in θ}{g} = 22 s$
Range (along north) $= \frac{u ^{2} s in 2 θ}{g} = 40 m$
Range (along east) $= \frac{1}{2} a T^{2} = 30 m$
$∴$ Range $= 3 0^{2} + 4 0^{2} = 50 m$