Question

a particle is projected horizontally with a speed of 20/root3 from some height at t equal to 0 . at what time will the velocity will make 60 angle with the initial velocity.

​

Answer 1

A particle is projected horizontal and the initial horizontal component of the velocity of the projected particle  

V $V_{h}$=20$\sqrt{3}$ , This will remain unchanged during its flight.

The initial vertical component of the velocity of the projected particle  

V $V_{v}$=0. This will change with time during its flight by the action of gravitational pull at

g =10$\frac{m}{s^{2} }$

Let after t  sec from start the particle's velocity velocity makes angle  θ = 60°

with the initial velocity in horizontal direction.

After  t sec from start its vertical component will be

$V'_{v}$=$V_{v}$+tx g=0+10t= 10t $\frac{m}{s}$

So $\frac{V'_{v} }{V_{h} }$ = tan θ

⇒ $\frac{10t}{20\sqrt{3} }$ = tan 60 °= $\sqrt{3}$

⇒ t = 2 sec

Answer 2

Answer:

2.04s

Explanation:

Vx=20√3,Vy=0

Vy'=Vy+gt

=0+9.8(T)

=9.8T

Tan 60°=Vy'/Vx

=9.8T/(20/√3)

T=Tan60×20/9.8×√3

=√3×20/9.8×√3

√3 gets cancelled

20/9.8=2.04s

In place of g=9.8 you can take g=10 also

then u will get answer as 2s