Question

A particle is projected horizontally with a speed of 20 by root 3 metre per second from some height at t is equal to zero at what time will its velocity makes 60 degree angle with the initial velocity​

Answer 1

Answer:

At time = 2.04 sec the particle's velocity will make an angle 60 degree angle with initial horizontal velocity.

Explanation:

A particle is thrown horizontally at a uniform initial velocity,  

vx = 20/√3 m/s

here, the vertical component of the velocity, vy = 0

initial time, t = 0 sec

Also, after some time “T” from the initial start, the vertical component of velocity of the particle makes an angle, θ = 60° with initial horizontal velocity.

Therefore, after “T” sec the vy’ will be  

vy’ = vy + (g * T)

or, vy’ = 0 + (9.8 * T) …….. [ taking g=9.8m/s²]

or, vy’ = 9.8T m/s

We know that, tan θ = y/x

∴ tan 60° = (vertical component of velocity, vy’) / (horizontal component of initial velocity, vx)

∴ tan 60° = (9.8 T) / (20/√3)

Or, T = (tan 60° * 20) / ( √3 * 9.8)

Or, T = (√3 * 200) / ( √3 * 9.8) ……… [ ∵ tan 60° = √3]

Or, T = 2.04 sec

Answer 2

Answer:

hope u understand it

Explanation:

At time velocity is v=vxi-gtj.

This makes an angle of 60 degree with vxi. Therefore ,

vxi.v=(vx)^2=vx (vx^2+g^2t^2)^1/2(1/2).

Squaring on both sides, after transferring 2 on left side, we have, 4(vx)^2=vx^2+g^2t^2. Then,

3(vx)^2=g^2t^2 or

t=(vx/g) sqrt of 3.