Question

A particle is projected with a velocity of 8 metre per second at an angle 45° with the horizontal what is the radius of curvature of the trajectory of the particle at the instant of One by fourth of the time of ascent is?​

Answer 1

Answer:

The radius of curvature of the trajectory of the particle at one fourth of ascent time is 7.6 metres.

Explanation:

Given Here,

Velocity of particle, v= 8 m/s

Angle of projection, O= 45 degrees

Time of travel, t=?

We know,

t= 2vsinO/g

or, t= (2×8×sin45)/9.81

or, t= 1.15s

Let, radius of the curvature and (t/4) or, T= y

So, y= -(gT2)/2 + vtanO

or, y= -[9.81×(0.29)2]/2 + 8tan45

or, y= 7.6m

The radius of curvature of the trajectory of the particle at one fourth of the descent time is 7.6 metres.