Question

plz help me with 166 question...??​

Answer 1

AnswEr :

Option (1) is correct

From the Question,

• Initial Velocity (u) = 20/√3 m/s

• Angle of Projection (∅) = 60°

Firstly,we would need to find the instant of time at which the velocity vector makes 60° with the horizontal

Time of Flight is given as :

$\sf \:T = \dfrac{2u sin(\alpha)}{g}$

Putting the values,

$\longrightarrow \: \sf \: T = \dfrac{2 \times \frac{ \: \: \: 20 \: }{ \sqrt{3} } \times \sin(60) }{10} \\ \\ \longrightarrow \: \sf \: T = \dfrac{2 \times 20 \times \sqrt{3} }{2 \times 10 \times \sqrt{3} } \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: T = 2s}}$

Velocity of a particle along the horizontal is ucos(∅) and is constant

We know that,

$\sf \: speed = \dfrac{distance }{time} \\ \\ \dashrightarrow \: \sf \: u. \cos( \alpha ) = \frac{x}{t} \\ \\ \dashrightarrow \: \sf \: x = \dfrac{20}{ \sqrt{3} } \times 2 \times \dfrac{1}{2} \\ \\ \dashrightarrow \ \boxed{ \boxed{\sf x = \dfrac{20}{ \sqrt{3} \: }m }}$

Distance travelled by the particle is 20/√3 m