Question

A particle is projected at an angle of 60° with speed of 10m/s. The time after which speed remains 1/√3 times of its initial speed is

Answer 1

Given initial speed u = 10 m/s

Angle of projection \(\theta = 60° \)

\(u_x = u\cos\theta\)

\(u_x = 10 \times cos60° \)

\(u_x = 10 \times \frac 12\)

\(u_x = 5 \,m/s\)

Speed will become \(\frac 1{\sqrt 3}\) of initial at t = \(\frac T{\sqrt 3}\) s, v_y = 0

\(t =\frac T{\sqrt 3}\)

\(t = \frac 1{\sqrt 3} \times \frac{2u\sin\theta}g\)

\(t = \frac 1{\sqrt 3} \times \frac{2\times 5 \times \sin60°}{10}\)

\(t =\frac 1{\sqrt 3 } \times \frac{10\times \sqrt 3}{10\times2}\)

\(t = \frac 12 sec\)